# NCERT Solutions Class 10 Maths Chapter 2

## NCERT Solutions for Class 10 Mathematics Chapter 2 Polynomials

Chapter 2 Class 10 Mathematics includes a detailed study of Polynomials. This chapter involves descriptions and practices of different equations and respective components. NCERT Class 10 Mathematics Chapter 2 textbook is one of the best study materials to understand the concept of Polynomials. The book has practice questions at the end of every chapter so that students can revise the concepts they have learned and prepare better for their exams.

Extramarks provides NCERT Solutions for Class 10 Mathematics Chapter 2 that have answers to all the questions given in the textbook. Whether students are looking for answers to questions or want to cross-check their answers, the NCERT Solutions by Extramarks are reliable learning material.

NCERT Solutions for Class 10 Maths Chapter 2 Polynomials –

### NCERT Solutions for Class 10 Mathematics

In addition to NCERT Solutions for Class 10 Mathematics Chapter 2, Extramarks also offers NCERT Solutions for all chapters in Class 10 Mathematics.

• Chapter 1 – Real Numbers
• Chapter 2 – Polynomials
• Chapter 3 – Pair of Linear Equations in Two Variables
• Chapter 4 – Quadratic Equations
• Chapter 5 – Arithmetic Progressions
• Chapter 6 – Triangles
• Chapter 7 – Coordinate Geometry
• Chapter 8 – Introduction to Trigonometry
• Chapter 9 – Some Applications of Trigonometry
• Chapter 10 – Circles
• Chapter 11 – Constructions
• Chapter 12 – Areas Related to Circles
• Chapter 13 – Surface Areas and Volumes
• Chapter 14 – Statistics
• Chapter 15 – Probability

### Polynomials: NCERT Solutions for Class 10 Mathematics Chapter 2 Summary

Chapter 2 includes the concepts of polynomials’ geometric and meanings of the terms relevant to it. The questions at the end of the NCERT Chapter 2 textbook will analyse the problem-solving skills of students. Therefore, Extramarks has ensured that every answer in solutions is solved in a step-by-step manner. The subject matter experts have kept the language simple and all the answers are highly accurate.

### Benefits of Using NCERT Solutions for Class 10 Mathematics Polynomials

There are many benefits of using NCERT Solutions for Class 10 Mathematics Chapter 2 polynomials. Let’s take a look at a few of them:

NCERT Solutions for Class 10 Mathematics Chapter 2 are created by subject matter experts. You can rest assured that all the answers have been framed by keeping in mind the guidelines provided by CBSE. Hence students who practise according to NCERT Solutions Class 10 Mathematics Chapter 2 will perform very well in their examinations.

• Clear Concept of Chapter 2 Class 10 Mathematics

Just knowing the answer to the question does not mean you excel in that subject, but understanding the path to find the solution should be the main aim of the student. Concept clarity helps students solve any problem with ease. As discussed above, NCERT Solutions for Class 10 Mathematics Chapter 2 focus more on sorting out conceptual queries of the students, thus helping them strengthen their subject knowledge for future standards.

• Clearing Doubts

It is natural for students to have considerable doubts while practising the questions of Class 10 Mathematics Chapter 2. These doubts can be quickly resolved with NCERT Solutions for Class 10 Mathematics Chapter 2

• Convenient Revision

NCERT Solutions for Class 10 Mathematics Chapter 2 make revision pretty convenient for students. They can refer to these solutions for last-minute preparations as well.

Related Questions

Q1. Write the following in the product form: a2b5.

Ans. a2b5 is – a×a×b×b×b×b×b

Q.1 The graphs of y = p(x) are given in figures below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.      Ans.

(i) The number of zeroes is 0 as the given graph does not intersect the x-axis.
(ii) The number of zeroes is 1 as the given graph intersects the x-axis at 1 point only.
(iii) The number of zeroes is 3 as the given graph intersects the x-axis at 3 points only.
(iv) The number of zeroes is 2 as the given graph intersects the x-axis at 2 points only.
(v) The number of zeroes is 4 as the given graph intersects the x-axis at 4 points only.
(vi) The number of zeroes is 3 as the given graph intersects the x-axis at 3 points only.

Q.2

$\begin{array}{l}\text{Find the zeroes of the following quadratic polynomials}\\ \text{and verify the relationship between the zeroes and}\\ \text{the coefficients.}\\ \text{(i)}{\mathrm{x}}^{2}-2\mathrm{x}-8\text{(ii)}4{\mathrm{s}}^{2}-4\mathrm{s}+1\text{(iii)}6{\mathrm{x}}^{2}-3-7\mathrm{x}\\ \text{(iv)}4{\mathrm{u}}^{2}+8\mathrm{u}\text{(v)}{\mathrm{t}}^{2}-15\text{(vi)}3{\mathrm{x}}^{2}-\mathrm{x}-4\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{}{\mathrm{x}}^{2}-2\mathrm{x}-8={\mathrm{x}}^{2}-4\mathrm{x}+2\mathrm{x}-8\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{x}\left(\mathrm{x}-4\right)+2\left(\mathrm{x}-4\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\mathrm{x}+2\right)\left(\mathrm{x}-4\right)\\ \text{So the value of}{\mathrm{x}}^{2}-2\mathrm{x}-8\text{is zero when}\mathrm{x}=-2\text{or}\mathrm{x}=4.\\ \text{Therefore, the zeroes of}{\mathrm{x}}^{2}-2\mathrm{x}-8\text{are}-2\text{and 4.}\\ \text{Now,}\\ \text{Sum of zeroes =}-2+4=2=\frac{-\left(-2\right)}{1}=\frac{-\left(\text{Coefficient of x)}}{{\text{Coefficient of x}}^{2}}\\ \text{Product of zeroes =}-2×4=\frac{-8}{1}=\frac{\text{Constant term}}{{\text{Coefficient of s}}^{2}}\\ \text{(ii)}\\ \text{}4{\mathrm{s}}^{2}-4\mathrm{s}+1={\left(2\mathrm{s}-1\right)}^{2}\\ \text{So the value of}4{\mathrm{s}}^{2}-4\mathrm{s}+1\text{is zero when}\mathrm{x}=\frac{1}{2}.\\ \text{Therefore, the zeroes of}4{\mathrm{s}}^{2}-4\mathrm{s}+1\text{are}\frac{1}{2}\text{and}\frac{1}{2}\text{.}\\ \text{Now,}\\ \text{Sum of zeroes =}\frac{1}{2}+\frac{1}{2}=1=\frac{-\left(-4\right)}{4}=\frac{-\left(\text{Coefficient of s)}}{{\text{Coefficient of s}}^{2}}\\ \text{Product of zeroes =}\frac{1}{2}×\frac{1}{2}=\frac{1}{4}=\frac{\text{Constant term}}{{\text{Coefficient of s}}^{2}}\\ \text{(iii)}\\ 6{\mathrm{x}}^{2}-3-7\mathrm{x}=6{\mathrm{x}}^{2}-9\mathrm{x}+2\mathrm{x}-3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{x}\left(2\mathrm{x}-3\right)+1\left(2\mathrm{x}-3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(3\mathrm{x}+1\right)\left(2\mathrm{x}-3\right)\\ \text{So the value of}6{\mathrm{x}}^{2}-3-7\mathrm{x}\text{is zero when}\mathrm{x}=\frac{-1}{3}\text{or}\mathrm{x}=\frac{3}{2}.\\ \text{Therefore, the zeroes of}6{\mathrm{x}}^{2}-3-7\mathrm{x}\text{are}\frac{-1}{3}\text{and}\frac{3}{2}\text{.}\\ \text{Now,}\\ \text{Sum of zeroes =}\frac{-1}{3}+\frac{3}{2}=\frac{7}{6}=\frac{-\left(-7\right)}{6}=\frac{-\left(\text{Coefficient of x)}}{{\text{Coefficient of x}}^{2}}\\ \text{Product of zeroes =}\frac{-1}{3}×\frac{3}{2}=\frac{-1}{2}=\frac{\text{Constant term}}{{\text{Coefficient of x}}^{2}}\\ \text{(iv)}\\ 4{\mathrm{u}}^{2}+8\mathrm{u}=4\mathrm{u}\left(\mathrm{u}+2\right)\\ \text{So the value of}4\mathrm{u}\left(\mathrm{u}+2\right)\text{is zero when}\mathrm{u}=0\text{or}\mathrm{x}=-2.\\ \text{Therefore, the zeroes of}4\mathrm{u}\left(\mathrm{u}+2\right)\text{are}0\text{and}-2.\\ \text{Now,}\\ \text{Sum of zeroes =}0-2=\frac{-8}{4}=\frac{-\left(\text{Coefficient of u)}}{{\text{Coefficient of u}}^{2}}\\ \text{Product of zeroes =}0×\left(-2\right)=0=\frac{0}{4}=\frac{\text{Constant term}}{{\text{Coefficient of u}}^{2}}\\ \text{(v)}\\ \text{\hspace{0.17em}}{\mathrm{t}}^{2}-15=\text{}{\mathrm{t}}^{2}-{\left(\sqrt{15}\right)}^{2}=\left(\mathrm{t}-\sqrt{15}\right)\left(\mathrm{t}+\sqrt{15}\right)\\ \text{So the value of}{\mathrm{t}}^{2}-15\text{is zero when}\mathrm{t}=\sqrt{15}\text{or}\mathrm{x}=-\sqrt{15}.\\ \text{Therefore, the zeroes of}{\mathrm{t}}^{2}-15\text{are}\sqrt{15}\text{and}-\sqrt{15}\text{.}\\ \text{Now,}\\ \text{Sum of zeroes =}\sqrt{15}-\sqrt{15}=0=\frac{0}{1}=\frac{-\left(\text{Coefficient of t)}}{{\text{Coefficient of t}}^{2}}\\ \text{Product of zeroes =}\sqrt{15}×\left(-\sqrt{15}\right)=-15=\frac{-15}{1}=\frac{\text{Constant term}}{{\text{Coefficient of t}}^{2}}\\ \text{(vi)}\\ \text{}3{\mathrm{x}}^{2}-\mathrm{x}-4=3{\mathrm{x}}^{2}-4\mathrm{x}+3\mathrm{x}-4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{x}\left(3\mathrm{x}-4\right)+1\left(3\mathrm{x}-4\right)=\left(3\mathrm{x}-4\right)\left(\mathrm{x}+1\right)\\ \text{So the value of}3{\mathrm{x}}^{2}-\mathrm{x}-4\text{is zero when}\mathrm{x}=\frac{4}{3}\text{or}\mathrm{x}=-1.\\ \text{Therefore, the zeroes of}3{\mathrm{x}}^{2}-\mathrm{x}-4\text{are}\frac{4}{3}\text{and}-1.\\ \text{Now,}\\ \text{Sum of zeroes =}\frac{4}{3}-1=\frac{1}{3}=\frac{-\left(-1\right)}{3}=\frac{-\left(\text{Coefficient of x)}}{{\text{Coefficient of x}}^{2}}\\ \text{Product of zeroes =}\frac{4}{3}×\left(-1\right)=\frac{-4}{3}=\frac{\text{Constant term}}{{\text{Coefficient of x}}^{2}}\end{array}$

Q.3

$\begin{array}{l}\text{Find a quadratic polynomial each with the given}\\ \text{numbers as the sum and product of its zeroes}\\ \text{respectively}.\\ \text{(i)}\frac{1}{4},\text{}-1\text{(ii)}\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{3}\text{(iii) 0,}\sqrt{5}\\ \text{(iv) 1, 1 (v)}\frac{-1}{4},\text{}\frac{1}{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(vi) 4, 1}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\frac{1}{4},\text{}-1\\ {\text{Let the quadratic polynomial be ax}}^{2}+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=\frac{1}{4}=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=-1=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{and thus}\\ \mathrm{a}=4,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=-1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=-4.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is 4x}}^{2}-\mathrm{x}-4.\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{3}\text{}\\ {\text{Let the quadratic polynomial be ax}}^{2}+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=\sqrt{2}=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=\frac{1}{3}=\frac{\mathrm{c}}{\mathrm{a}}\\ \text{Now,}\\ \mathrm{\alpha }+\mathrm{\beta }=\sqrt{2}=\frac{3\sqrt{2}}{3}=\frac{-\mathrm{b}}{\mathrm{a}}\text{}\\ \text{and}\mathrm{\alpha \beta }=\frac{1}{3}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=3,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=-3\sqrt{2},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=1.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is 3x}}^{2}-3\sqrt{2}\mathrm{x}+1.\\ \text{(iii) 0,}\sqrt{5}\\ {\text{Let the quadratic polynomial be ax}}^{2}+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=0=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=\sqrt{5}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=0,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=\sqrt{5}.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is x}}^{2}+\sqrt{5}.\\ \text{(iv) 1, 1}\\ {\text{Let the quadratic polynomial be ax}}^{2}+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=1=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=1=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=-1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=1.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is x}}^{2}-\mathrm{x}+1.\\ \\ \text{(v)}\frac{-1}{4},\text{}\frac{1}{4}\text{}\\ {\text{Let the quadratic polynomial be ax}}^{2}+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=\frac{-1}{4}=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=\frac{1}{4}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=4,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=1.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is 4x}}^{2}+\mathrm{x}+1.\\ \text{(vi) 4, 1}\\ {\text{Let the quadratic polynomial be ax}}^{2}+\mathrm{bx}+\mathrm{c}\text{and its}\\ \text{zeroes be}\mathrm{\alpha }\text{and}\mathrm{\beta }\text{. Then, we have}\\ \mathrm{\alpha }+\mathrm{\beta }=4=\frac{-\mathrm{b}}{\mathrm{a}}\text{and}\mathrm{\alpha \beta }=1=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=-4,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=1.\\ \text{So, the quadratic polynomial which fits the given}\\ {\text{conditions is x}}^{2}-4\mathrm{x}+1.\end{array}$

Q.4

$\begin{array}{l}\text{Divide the polynomial}\mathrm{p}\left(\mathrm{x}\right)\text{by the polynomial}\mathrm{g}\left(\mathrm{x}\right)\\ \text{and find the quotient and remainder in each of the}\\ \text{following:}\\ \text{(i)}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+5\mathrm{x}-3,\text{}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^{2}-2\\ \text{(ii)}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^{4}-3{\mathrm{x}}^{2}+4\mathrm{x}+5,\text{}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^{2}+1-\mathrm{x}\\ \text{(iii)}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^{4}-5\mathrm{x}+6,\text{}\mathrm{g}\left(\mathrm{x}\right)=2-{\mathrm{x}}^{2}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\text{}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+5\mathrm{x}-3,\text{}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^{2}-2\\ \\ {\mathrm{x}}^{2}-2\begin{array}{c}\mathrm{x}-3\\ \overline{){\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+5\mathrm{x}-3}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\underset{¯}{\begin{array}{l}{\mathrm{x}}^{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-2\mathrm{x}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}-3{\mathrm{x}}^{2}+7\mathrm{x}-3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-3{\mathrm{x}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+6\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-}{7\mathrm{x}-9}\\ \text{Quotient}=\mathrm{x}-3\\ \text{Remainder}=7\mathrm{x}-9\\ \text{(ii)}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^{4}-3{\mathrm{x}}^{2}+4\mathrm{x}+5,\text{}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^{2}+1-\mathrm{x}\\ \\ {\mathrm{x}}^{2}+1-\mathrm{x}\begin{array}{c}{\mathrm{x}}^{2}+\mathrm{x}-3\\ \overline{){\mathrm{x}}^{4}-3{\mathrm{x}}^{2}+4\mathrm{x}+5}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}{\mathrm{x}}^{4}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-{\mathrm{x}}^{3}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{3}-4{\mathrm{x}}^{2}+4\mathrm{x}+5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{3}-{\mathrm{x}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-3{\mathrm{x}}^{2}+3\mathrm{x}+5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}-3{\mathrm{x}}^{2}+3\mathrm{x}-3\\ +\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8\\ \text{Quotient}={\mathrm{x}}^{2}+\mathrm{x}-3\\ \text{Remainder}=8\\ \text{(iii)}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^{4}-5\mathrm{x}+6,\text{}\mathrm{g}\left(\mathrm{x}\right)=2-{\mathrm{x}}^{2}\\ \\ 2-{\mathrm{x}}^{2}\begin{array}{c}-{\mathrm{x}}^{2}-2\\ \overline{){\mathrm{x}}^{4}-5\mathrm{x}+6}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}{\mathrm{x}}^{4}-2\text{\hspace{0.17em}}{\mathrm{x}}^{2}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}2{\mathrm{x}}^{2}-5\mathrm{x}+6\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}\text{\hspace{0.17em}}2{\mathrm{x}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-5\mathrm{x}+10\\ \text{Quotient}=-{\mathrm{x}}^{2}-2\\ \text{Remainder}=-5\mathrm{x}+10\end{array}$

Q.5

$\begin{array}{l}\text{Check whether the first polynomial is a facator of}\\ \text{the second polymomial by dividing the second}\\ \text{polynomial by the first polynomial:}\\ \text{(i)\hspace{0.17em}\hspace{0.17em}}{\mathrm{t}}^{2}-3,\text{2}{\mathrm{t}}^{4}+3{\mathrm{t}}^{3}-2{\mathrm{t}}^{2}-9\mathrm{t}-12\\ \text{(ii)}{\mathrm{x}}^{2}+3\mathrm{x}+1,\text{3}{\mathrm{x}}^{4}+5{\mathrm{x}}^{3}-7{\mathrm{x}}^{2}+2\mathrm{x}+2\\ \text{(iii)}{\mathrm{x}}^{3}-3\mathrm{x}+1,{\text{x}}^{5}-4{\mathrm{x}}^{3}+{\mathrm{x}}^{2}+3\mathrm{x}+1\text{}\end{array}$

Ans.

$\begin{array}{l}\text{(i)\hspace{0.17em}\hspace{0.17em}}{\mathrm{t}}^{2}-3,\text{2}{\mathrm{t}}^{4}+3{\mathrm{t}}^{3}-2{\mathrm{t}}^{2}-9\mathrm{t}-12\\ \\ {\mathrm{t}}^{2}-3\begin{array}{c}2{\mathrm{t}}^{2}+3\mathrm{t}+4\\ \overline{)\text{2}{\mathrm{t}}^{4}+3{\mathrm{t}}^{3}-2{\mathrm{t}}^{2}-9\mathrm{t}-12}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}2{\mathrm{t}}^{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-6{\mathrm{t}}^{2}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3{\mathrm{t}}^{3}+4{\mathrm{t}}^{2}-9\mathrm{t}-12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}3{\mathrm{t}}^{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-9\mathrm{t}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4{\mathrm{t}}^{2}-12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4{\mathrm{t}}^{2}-12\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0\\ \text{Remainder = 0}\\ \text{So,}{\mathrm{t}}^{2}-3\text{is a factor of 2}{\mathrm{t}}^{4}+3{\mathrm{t}}^{3}-2{\mathrm{t}}^{2}-9\mathrm{t}-12.\\ \\ \text{(ii)}{\mathrm{x}}^{2}+3\mathrm{x}+1,\text{3}{\mathrm{x}}^{4}+5{\mathrm{x}}^{3}-7{\mathrm{x}}^{2}+2\mathrm{x}+2\\ \\ {\mathrm{x}}^{2}+3\mathrm{x}+1\begin{array}{c}3{\mathrm{x}}^{2}-4\mathrm{x}+2\\ \overline{)\text{3}{\mathrm{x}}^{4}+5{\mathrm{x}}^{3}-7{\mathrm{x}}^{2}+2\mathrm{x}+2}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}3{\mathrm{x}}^{4}+9{\mathrm{x}}^{3}+3{\mathrm{x}}^{2}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-4{\mathrm{x}}^{3}-10{\mathrm{x}}^{2}+2\mathrm{x}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}\text{\hspace{0.17em}}-4{\mathrm{x}}^{3}\text{\hspace{0.17em}}-12{\mathrm{x}}^{2}-4\mathrm{x}\\ \text{\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2{\mathrm{x}}^{2}+6\mathrm{x}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2{\mathrm{x}}^{2}+6\mathrm{x}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0\\ \text{Remainder = 0}\\ \text{So,}{\mathrm{x}}^{2}+3\mathrm{x}+1\text{is a factor of 3}{\mathrm{x}}^{4}+5{\mathrm{x}}^{3}-7{\mathrm{x}}^{2}+2\mathrm{x}+2.\\ \\ \text{(iii)}{\mathrm{x}}^{3}-3\mathrm{x}+1,{\text{x}}^{5}-4{\mathrm{x}}^{3}+{\mathrm{x}}^{2}+3\mathrm{x}+1\\ \\ {\mathrm{x}}^{3}-3\mathrm{x}+1\begin{array}{c}{\mathrm{x}}^{2}-1\\ \overline{){\text{x}}^{5}-4{\mathrm{x}}^{3}+{\mathrm{x}}^{2}+3\mathrm{x}+1}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}{\mathrm{x}}^{5}-3{\mathrm{x}}^{3}+{\mathrm{x}}^{2}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}-{\mathrm{x}}^{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+3\mathrm{x}+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}-{\mathrm{x}}^{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+3\mathrm{x}-1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\\ \text{Remainder = 2}\\ \text{So,}{\mathrm{x}}^{3}-3\mathrm{x}+1{\text{is not a factor of x}}^{5}-4{\mathrm{x}}^{3}+{\mathrm{x}}^{2}+3\mathrm{x}+1.\end{array}$

Q.6

$\begin{array}{l}\text{Obtain all other zeroes of 3}{\mathrm{x}}^{4}+6{\mathrm{x}}^{3}-2{\mathrm{x}}^{2}-10\mathrm{x}-5,\\ \text{if two of its zeroes are}\sqrt{\frac{5}{3}\text{}}\text{and}-\sqrt{\frac{5}{3}\text{}}.\end{array}$

Ans.

$\begin{array}{l}{\text{p(x)=3x}}^{\text{4}}{\text{+6x}}^{\text{3}}-{\text{2x}}^{\text{2}}-\text{10x-5}\\ \text{The two zeroes of p(x) are}\sqrt{\frac{\text{5}}{\text{3}}\text{}}\text{and-}\sqrt{\frac{\text{5}}{\text{3}}\text{}}\text{.}\\ \text{Therefore,}\left(\text{x}-\sqrt{\frac{\text{5}}{\text{3}}\text{}}\right)\text{and}\left(\text{x+}\sqrt{\frac{\text{5}}{\text{3}}\text{}}\right)\text{are factors of p(x).}\\ \text{Also,}\\ \left(\text{x}-\sqrt{\frac{\text{5}}{\text{3}}\text{}}\right)\left(\text{x+}\sqrt{\frac{\text{5}}{\text{3}}\text{}}\right){\text{ = x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\\ {\text{and so x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\text{is a factor of p(x).}\\ \text{Now,}\\ {\text{ 3x}}^{\text{2}}\text{+6x+3}\\ {\text{x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\text{ }\overline{){\text{3x}}^{\text{4}}{\text{+6x}}^{\text{3}}-{\text{2x}}^{\text{2}}-\text{10x}-\text{5}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}{\text{3x}}^{\text{4}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-{\text{5x}}^{\text{2}}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}6x}}^{\text{3}}{\text{+3x}}^{\text{2}}-\text{10x}-\text{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}{\text{6x}}^{\text{3}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{10x}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3x}}^{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}{\text{3x}}^{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{5}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}+}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0}\\ {\text{3x}}^{\text{4}}{\text{+6x}}^{\text{3}}-{\text{2x}}^{\text{2}}-\text{10x}-\text{5=}\left({\text{x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\right)\left({\text{3x}}^{\text{2}}\text{+6x+3}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3}\left({\text{x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\right)\left({\text{x}}^{\text{2}}\text{+2x+1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=3}\left({\text{x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\right)\left(\text{x+1}\right)\text{(x+1)}\\ \text{Equating}\left({\text{x}}^{\text{2}}-\frac{\text{5}}{\text{3}}\right)\left(\text{x+1}\right)\text{(x+1) equal to zero, we get}\\ \text{the zeroes of the given polynomial.}\\ \text{Hence, the zeroes of the given polynomial are}\sqrt{\frac{\text{5}}{\text{3}}\text{}}\text{,\hspace{0.17em}}-\sqrt{\frac{\text{5}}{\text{3}}\text{}}\text{,\hspace{0.17em} 1}\\ \text{and }-\text{1.}\end{array}$

Q.7

$\begin{array}{l}\text{On dividing }{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+2\text{ by a polynomial}\mathrm{g}\left(\mathrm{x}\right),\\ \text{the quotient and remainder were }\mathrm{x}-2\text{ and }-2\mathrm{x}+4,\\ \text{respectively. Find }\mathrm{g}\left(\mathrm{x}\right)\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{Dividend =}{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+2\\ \text{Quotient =}\mathrm{x}-2\\ \text{Remainder =}-2\mathrm{x}+4\\ \text{Divisor = g(x) = ?}\\ \text{We know that,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividend}=\text{Divisor}×\text{Quotient}+\text{Remainder}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+2=\text{g(x)}×\left(\text{}\mathrm{x}-2\right)+\left(-2\mathrm{x}+4\right)\\ ⇒\text{g(x)}×\left(\text{}\mathrm{x}-2\right)={\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+2+2\mathrm{x}-4\\ ⇒\text{g(x)}=\frac{\left({\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+3\mathrm{x}-2\right)}{\left(\text{}\mathrm{x}-2\right)}\\ \\ \mathrm{x}-2\begin{array}{c}{\mathrm{x}}^{2}-\mathrm{x}+1\\ \overline{){\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+3\mathrm{x}-2}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\underset{¯}{\begin{array}{l}{\mathrm{x}}^{3}-2{\mathrm{x}}^{2}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-{\mathrm{x}}^{2}+3\mathrm{x}-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-{\mathrm{x}}^{2}+2\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^{2}-\mathrm{x}+1\end{array}$

Q.8

$\begin{array}{l}\text{Give examples of polynomials}\mathrm{p}\left(\mathrm{x}\right),\mathrm{g}\left(\mathrm{x}\right),\mathrm{q}\left(\mathrm{x}\right)\text{and}\mathrm{r}\left(\mathrm{x}\right)\\ \text{which satisfy the division algorithm and}\\ \text{(i) deg}\mathrm{p}\left(\mathrm{x}\right)=\text{deg}\mathrm{q}\left(\mathrm{x}\right)\text{(ii) deg}\mathrm{q}\left(\mathrm{x}\right)=\text{deg}\mathrm{r}\left(\mathrm{x}\right)\\ \text{(iii) deg}\mathrm{r}\left(\mathrm{x}\right)=0\end{array}$

Ans.

$\begin{array}{l}\text{(i) deg}\mathrm{p}\left(\mathrm{x}\right)=\text{deg}\mathrm{q}\left(\mathrm{x}\right)\\ \text{if divisor is constant then}\\ \text{Degree of quotient = degree of dividend}\\ \text{Let dividend}\mathrm{p}\left(\mathrm{x}\right)=2{\mathrm{x}}^{2}-2\mathrm{x}+14\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}divisor \hspace{0.17em}}\mathrm{g}\left(\mathrm{x}\right)=2\\ \text{Then, wer have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}quotient\hspace{0.17em}\hspace{0.17em}}\mathrm{q}\left(\mathrm{x}\right)={\mathrm{x}}^{2}-\mathrm{x}+7\\ \text{and remainder\hspace{0.17em}\hspace{0.17em}}\mathrm{r}\left(\mathrm{x}\right)=0\\ \text{We find that}\\ \text{deg}\mathrm{p}\left(\mathrm{x}\right)=\text{deg}\mathrm{q}\left(\mathrm{x}\right)=2\\ \text{Let us check for division algorithm.}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}dividend}=\text{divisor}×\text{quotient + remainder}\\ \text{or}2{\mathrm{x}}^{2}-2\mathrm{x}+14=2\left({\mathrm{x}}^{2}-\mathrm{x}+7\right)+0=\text{}2{\mathrm{x}}^{2}-2\mathrm{x}+14\\ \text{Thus, the division algorithm is satisfied.}\\ \text{(ii) deg}\mathrm{q}\left(\mathrm{x}\right)=\text{deg}\mathrm{r}\left(\mathrm{x}\right)\\ \\ \text{Let us divide}{\mathrm{x}}^{3}+\mathrm{x}\text{by}{\mathrm{x}}^{2}.\text{}\\ \text{Clearly we have,}\\ \text{dividend}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^{3}+\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}divisor \hspace{0.17em}}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}quotient\hspace{0.17em}\hspace{0.17em}}\mathrm{q}\left(\mathrm{x}\right)=\mathrm{x}\\ \text{and remainder\hspace{0.17em}\hspace{0.17em}}\mathrm{r}\left(\mathrm{x}\right)=\mathrm{x}\\ \text{Here,}\\ \text{deg}\mathrm{q}\left(\mathrm{x}\right)=\text{deg}\mathrm{r}\left(\mathrm{x}\right)=1\\ \text{Let us check for division algorithm.}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}dividend}=\text{divisor}×\text{quotient + remainder}\\ \text{or}{\mathrm{x}}^{3}+\mathrm{x}={\mathrm{x}}^{2}\cdot \mathrm{x}+\mathrm{x}={\mathrm{x}}^{3}+\mathrm{x}\text{}\\ \text{Thus, the division algorithm is satisfied.}\\ \text{(iii) deg}\mathrm{r}\left(\mathrm{x}\right)=0\\ \\ \text{Degree of remainder will be zero if the remainder is constant.}\\ \text{Let us divide}{\mathrm{x}}^{2}+1\text{by}\mathrm{x}.\text{}\\ \text{Clearly we have,}\\ \text{dividend}\mathrm{p}\left(\mathrm{x}\right)={\mathrm{x}}^{2}+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}divisor \hspace{0.17em}}\mathrm{g}\left(\mathrm{x}\right)=\mathrm{x}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}quotient\hspace{0.17em}\hspace{0.17em}}\mathrm{q}\left(\mathrm{x}\right)=\mathrm{x}\\ \text{and remainder\hspace{0.17em}\hspace{0.17em}}\mathrm{r}\left(\mathrm{x}\right)=1\\ \text{Here,}\\ \text{deg}\mathrm{r}\left(\mathrm{x}\right)=0\\ \text{Let us check for division algorithm.}\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}dividend}=\text{divisor}×\text{quotient + remainder}\\ \text{or}{\mathrm{x}}^{2}+1=\mathrm{x}\cdot \mathrm{x}+1={\mathrm{x}}^{2}+1\text{}\\ \text{Thus, the division algorithm is satisfied.}\end{array}$

Q.9

$\begin{array}{l}\text{Verify that the numbers given alongside of the cubic}\\ \text{polynomials below are their zeroes. Also verify the}\\ \text{relationship between the zeroes and the coefficients}\\ \text{in each case:}\\ {\text{(i) 2x}}^{3}+{\mathrm{x}}^{2}-5\mathrm{x}+2;\text{}\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-2\text{\hspace{0.17em}}\\ {\text{(ii) x}}^{3}-4{\mathrm{x}}^{2}+5\mathrm{x}-2;\text{2, 1, 1}\end{array}$

Ans.

$\begin{array}{l}{\text{(i) 2x}}^{3}+{\mathrm{x}}^{2}-5\mathrm{x}+2;\text{}\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-2\text{\hspace{0.17em}}\\ \text{Let}\mathrm{p}\left(\mathrm{x}\right)={\text{2x}}^{3}+{\mathrm{x}}^{2}-5\mathrm{x}+2\\ \text{Then}\\ \mathrm{p}\left(\frac{1}{2}\right)=\text{2}{\left(\frac{1}{2}\right)}^{3}+{\left(\frac{1}{2}\right)}^{2}-5\left(\frac{1}{2}\right)+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2=\frac{1}{2}+2-\frac{5}{2}=0\\ \\ \mathrm{p}\left(1\right)=\text{2}{\left(1\right)}^{3}+{\left(1\right)}^{2}-5\left(1\right)+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2+1-5+2=0\\ \mathrm{p}\left(-2\right)=\text{2}{\left(-2\right)}^{3}+{\left(-2\right)}^{2}-5\left(-2\right)+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-16+4+10+2=0\\ \\ \text{Therefore,}\frac{1}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1\text{and\hspace{0.17em}\hspace{0.17em}}-2\text{are zeroes of the given poynomial.}\\ \text{Comparing the given poynomial with}{\mathrm{ax}}^{3}+{\mathrm{bx}}^{2}+\mathrm{cx}+\mathrm{d},\text{we get}\\ \mathrm{a}=2,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=-5,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=2\\ \text{Let the given roots are}\mathrm{\alpha },\text{\hspace{0.17em}}\mathrm{\beta }\text{and}\mathrm{\gamma }\text{. Then}\\ \mathrm{\alpha }\text{\hspace{0.17em}=\hspace{0.17em}}\frac{1}{2},\text{}\mathrm{\beta }=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\gamma }=-2\\ \text{Now,}\\ \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=\frac{1}{2}+1-2=-\frac{1}{2}=\frac{-\mathrm{b}}{\mathrm{a}}\\ \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=\frac{1}{2}×1+1×\left(-2\right)+\left(-2\right)×\frac{1}{2}=\frac{-5}{2}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{\alpha \beta \gamma }=\frac{1}{2}×1×\left(-2\right)=-1=\frac{-2}{2}=\frac{-\mathrm{d}}{\mathrm{a}}\\ \\ \text{The relationship between roots and coefficients is verified.}\\ {\text{(ii) x}}^{3}-4{\mathrm{x}}^{2}+5\mathrm{x}-2;\text{2, 1, 1}\\ \text{Let}\mathrm{p}\left(\mathrm{x}\right)={\text{x}}^{3}-4{\mathrm{x}}^{2}+5\mathrm{x}-2\\ \text{Then}\\ \mathrm{p}\left(\text{2}\right)={\left(2\right)}^{3}-4{\left(2\right)}^{2}+5\left(2\right)-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8-16+10-2=0\\ \\ \mathrm{p}\left(1\right)={\left(1\right)}^{3}-4{\left(1\right)}^{2}+5\left(1\right)-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-4+5-2=0\\ \\ \text{Therefore,}2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1\text{and\hspace{0.17em}\hspace{0.17em}}1\text{are zeroes of the given poynomial.}\\ \text{Comparing the given poynomial with}{\mathrm{ax}}^{3}+{\mathrm{bx}}^{2}+\mathrm{cx}+\mathrm{d},\text{we get}\\ \mathrm{a}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=-4,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=5,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=-2\\ \text{Let the given roots are}\mathrm{\alpha },\text{\hspace{0.17em}}\mathrm{\beta }\text{and}\mathrm{\gamma }\text{. Then}\\ \mathrm{\alpha }\text{\hspace{0.17em}=\hspace{0.17em}}2,\text{}\mathrm{\beta }=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\gamma }=1\\ \text{Now,}\\ \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=2+1+1=4=-\frac{-4}{1}=\frac{-\mathrm{b}}{\mathrm{a}}\\ \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=2×1+1×1+1×2=5=\frac{5}{1}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{\alpha \beta \gamma }=2×1×1=2=\frac{-\left(-2\right)}{1}=\frac{-\mathrm{d}}{\mathrm{a}}\\ \text{The relationship between roots and coefficients is verified.}\end{array}$

Q.10

$\begin{array}{l}\text{Find a cubic polynomial with the sum of zeroes , sum of the}\\ \text{product of its zeroes taken two at a time, and the}\\ \text{product of its zeroes as 2,}-\text{7,}-\text{14 respectively.}\end{array}$

Ans.

$\begin{array}{l}\text{Let the polynomial be}{\mathrm{ax}}^{3}+{\mathrm{bx}}^{2}+\mathrm{cx}+\mathrm{d}\text{and its zeroes}\\ \text{be}\mathrm{\alpha }\text{,\hspace{0.17em}\hspace{0.17em}}\mathrm{\beta }\text{and}\mathrm{\gamma }.\\ \text{Given that}\\ \mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=2=\frac{-\left(-2\right)}{1}=\frac{-\mathrm{b}}{\mathrm{a}}\\ \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=-7=\frac{-7}{1}=\frac{\mathrm{c}}{\mathrm{a}}\\ \mathrm{\alpha \beta \gamma }=-14=\frac{-14}{1}=\frac{-\mathrm{d}}{\mathrm{a}}\\ \text{If we take}\mathrm{a}=1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=-2,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{c}=-7\text{and}\mathrm{d}=14\text{\hspace{0.17em}\hspace{0.17em}then the}\\ \text{required polynomial is}{\mathrm{x}}^{3}-2{\mathrm{x}}^{2}-7\mathrm{x}+14.\end{array}$

Q.11

$\begin{array}{l}\text{If the zeroes of the polynomial }{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+1\\ \text{are a}-\text{b, a, }\mathrm{a}+\mathrm{b}\text{, find a and b.}\end{array}$

Ans.

$\begin{array}{l}\text{The given polynomial is}{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+1\text{\hspace{0.17em}\hspace{0.17em}and its zeroes}\\ \text{are a}-\text{b, a,}\mathrm{a}+\mathrm{b}.\\ \text{Sum of the roots = (a}-\text{b}\right)+\mathrm{a}+\left(\mathrm{a}+\mathrm{b}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{a}=-\frac{\text{Coefficient of}{\mathrm{x}}^{2}}{\text{Coefficient of}{\mathrm{x}}^{3}}=3\\ ⇒\mathrm{a}=1\\ \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sum of the products of zero taken two at a time}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{a}\left(\mathrm{a}-\mathrm{b}\right)+\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right)+\left(\mathrm{a}+\mathrm{b}\right)\left(\mathrm{a}-\mathrm{b}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{a}}^{2}-\mathrm{ab}+{\mathrm{a}}^{2}+\mathrm{ab}+{\mathrm{a}}^{2}-{\mathrm{b}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3{\mathrm{a}}^{2}-{\mathrm{b}}^{2}=\frac{\text{Coefficient of}\mathrm{x}}{\text{Coefficient of}{\mathrm{x}}^{3}}=1\\ ⇒3{\mathrm{a}}^{2}-{\mathrm{b}}^{2}=1\\ ⇒3.1-{\mathrm{b}}^{2}=1\\ ⇒{\mathrm{b}}^{2}=2\\ ⇒\mathrm{b}=±\sqrt{2}\\ \text{Hence,}\mathrm{a}=1\text{and}\mathrm{b}=\sqrt{2}\text{\hspace{0.17em}\hspace{0.17em}or}-\sqrt{2}\end{array}$

Q.12

$\begin{array}{l}\text{If two zeroes of the polynomial}\text{ }{x}^{4}-6{x}^{3}-26{x}^{2}+138x-35\\ \text{are}\text{ }2±\sqrt{3}\text{,}\text{ }\text{find other zeroes}\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{Two zeroes of the polynomial}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35\\ \text{are}2±\sqrt{3}.\\ \text{Therefore,}\left[\mathrm{x}-\left(2+\sqrt{3}\right)\right]\text{and}\left[\mathrm{x}-\left(2-\sqrt{3}\right)\right]\text{are factors of the}\\ \text{given polynomial.}\\ \text{Now},\\ \left[\mathrm{x}-\left(2+\sqrt{3}\right)\right]\left[\mathrm{x}-\left(2-\sqrt{3}\right)\right]=\left[\left(\mathrm{x}-2\right)-\sqrt{3}\right)\right]\left[\left(\mathrm{x}-2\right)+\sqrt{3}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(\mathrm{x}-2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}-4\mathrm{x}+4-3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}-4\mathrm{x}+1\\ {\mathrm{x}}^{2}-4\mathrm{x}+1\text{is also a factor of the given polynomial.}\\ \text{Therefore, we divide}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35\\ \text{by}{\mathrm{x}}^{2}-4\mathrm{x}+1\text{to find other factors.}\\ \\ {\mathrm{x}}^{2}-4\mathrm{x}+1\overline{){\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{4}-4{\mathrm{x}}^{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+{\mathrm{x}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\overline{\begin{array}{l}-2{\mathrm{x}}^{3}-27{\mathrm{x}}^{2}+138\mathrm{x}-35\\ -2{\mathrm{x}}^{3}+8{\mathrm{x}}^{2}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}\\ +\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\overline{\begin{array}{l}-35{\mathrm{x}}^{2}+140\mathrm{x}-35\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ -35{\mathrm{x}}^{2}+140\mathrm{x}-35\\ +\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\overline{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\\ {\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left({\mathrm{x}}^{2}-4\mathrm{x}+1\right)\left({\mathrm{x}}^{2}-2\mathrm{x}-35\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left[\mathrm{x}-\left(2+\sqrt{3}\right)\right]\left[\mathrm{x}-\left(2-\sqrt{3}\right)\right]\left({\mathrm{x}}^{2}-7\mathrm{x}+5\mathrm{x}-35\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left[\mathrm{x}-\left(2+\sqrt{3}\right)\right]\left[\mathrm{x}-\left(2-\sqrt{3}\right)\right]\left(\mathrm{x}-7\right)\left(\mathrm{x}+5\right)\\ \left(\mathrm{x}-7\right)\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\left(\mathrm{x}+5\right)\text{\hspace{0.17em}}\mathrm{are}\text{\hspace{0.17em}}\mathrm{factors}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35.\\ \mathrm{Hence},\text{\hspace{0.17em}}7\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}-5\text{\hspace{0.17em}}\mathrm{are}\text{\hspace{0.17em}}\mathrm{zeroes}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}-26{\mathrm{x}}^{2}+138\mathrm{x}-35.\end{array}$

Q.13

$\begin{array}{l}\text{If the polynomial }{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+16{\mathrm{x}}^{2}-25\mathrm{x}+10\text{ is divided}\\ \text{by another polynomial }{\mathrm{x}}^{2}-2\mathrm{x}+\mathrm{k}\text{, the remainder comes}\\ \text{out to be }\mathrm{x}+\mathrm{a}\text{, find }\mathrm{k}\text{ and }\mathrm{a}\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{We know that}\\ \text{​​​​​\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Dividend = Divisor}×\text{Quotient + Remainder}\\ \text{or Dividend}-\text{Remainder = Divisor}×\text{Quotient}\\ \text{or}\left({\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+16{\mathrm{x}}^{2}-25\mathrm{x}+10\right)-\left(\mathrm{x}+\mathrm{a}\right)=\left({\mathrm{x}}^{2}-2\mathrm{x}+\mathrm{k}\right)×\text{Quotient}\\ \text{or}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+16{\mathrm{x}}^{2}-26\mathrm{x}+10-\mathrm{a}=\left({\mathrm{x}}^{2}-2\mathrm{x}+\mathrm{k}\right)×\text{Quotient}\\ \text{Now},\\ {\mathrm{x}}^{2}-2\mathrm{x}+\mathrm{k}\begin{array}{c}{\mathrm{x}}^{2}-4\mathrm{x}+\left(8-\mathrm{k}\right)\\ \overline{)\begin{array}{l}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+16{\mathrm{x}}^{2}-26\mathrm{x}+10-\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\end{array}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}{\mathrm{x}}^{4}-2{\mathrm{x}}^{3}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{kx}}^{2}\\ -\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-4{\mathrm{x}}^{3}+\left(16-\mathrm{k}\right){\mathrm{x}}^{2}-26\mathrm{x}+10-\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-4{\mathrm{x}}^{3}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8{\mathrm{x}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}}4\mathrm{kx}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\underset{¯}{\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\left(8-\mathrm{k}\right){\mathrm{x}}^{2}+\left(4\mathrm{k}-26\right)\mathrm{x}+10-\mathrm{a}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\left(8-\mathrm{k}\right){\mathrm{x}}^{2}-\left(-2\mathrm{k}+16\right)\mathrm{x}+8\mathrm{k}-{\mathrm{k}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}-\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(2\mathrm{k}-10\right)\mathrm{x}+{\mathrm{k}}^{2}-8\mathrm{k}+10-\mathrm{a}\\ \text{Remainder\hspace{0.17em}\hspace{0.17em}}\left(2\mathrm{k}-10\right)\mathrm{x}+{\mathrm{k}}^{2}-8\mathrm{k}+10-\mathrm{a}\text{must be zero.}\\ \mathrm{So},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(2\mathrm{k}-10\right)\mathrm{x}+{\mathrm{k}}^{2}-8\mathrm{k}+10-\mathrm{a}\text{}=0\\ ⇒2\mathrm{k}-10=0\text{or}{\mathrm{k}}^{2}-8\mathrm{k}+10-\mathrm{a}=0\\ ⇒\mathrm{k}=5\text{or}{5}^{2}-8×5+10-\mathrm{a}=0\\ ⇒\mathrm{k}=5\text{and}\mathrm{a}=-5\end{array}$

## 1. Why should you prefer using NCERT Solutions for Class 10 Mathematics Chapter 2 Polynomials?

Using these NCERT Solutions for Class 10 Mathematics Chapter 2 will enhance the problem-solving skills of students and strengthen their conceptual knowledge.

## 2. How to practise Class 10 Polynomial questions?

Students should practise the questions given in NCERT textbook and should refer to NCERT Solutions Class 10 Mathematics Chapter 2 in case they get stuck on a particular question

## 3. How many exercises does Chapter 2 Class 10 Mathematics consist of?

The Class 10 Chapter 2 Polynomials has four exercises. Extramarks provides Solutions for Chapter 2 for all four exercises.