NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials Exercise 2.3

NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials Exercise 2.3 are provided here to help students with their Class 10 exam preparations. These solutions are carefully designed by our subject experts to ensure students understand the roots of polynomials, factorization, and relation between coefficients and zeroes of a polynomial in a step-by-step manner.

Exercise 2.3 in Polynomials focuses on finding the zeroes of quadratic polynomials and expressing the factorized form of polynomials. This exercise strengthens students' understanding of relationship between the zeroes and coefficients of quadratic polynomials and how to use this relationship to solve problems effectively.

NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials Exercise 2.3

Polynomials

Medium

Q.

Find a quadratic polynomial with the given numbers as the sum and product of its zeroes respectively.

​$$(i) \dfrac{1}{4},{ }-1{          (ii) }\sqrt{2},{\hspace{0.17em}\hspace{0.17em}}\frac{1}{3}{          (iii) 0, }\sqrt{5}\\ {(iv) 1, 1             (v) }\frac{-1}{4},{ }\frac{1}{4}{      \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(vi) 4, 1}$$​​​

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Polynomials

Medium

Q.

$\begin{array}{l}{On dividing\hspace{0.33em}}{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+2{\hspace{0.33em}by a polynomial }\mathrm{g}\left(\mathrm{x}\right),\\ {the quotient and remainder were\hspace{0.33em}}\mathrm{x}-2{\hspace{0.33em}and\hspace{0.33em}}-2\mathrm{x}+4,\\ {respectively. Find\hspace{0.33em}}\mathrm{g}\left(\mathrm{x}\right){.}\end{array}$

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Polynomials

Medium

Q.

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

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Polynomials

Medium

Q.

$\begin{array}{l}{If the zeroes of the polynomial\hspace{0.33em}}{\mathrm{x}}^{3}-3{\mathrm{x}}^{2}+\mathrm{x}+1\\ {are a}-{b, a,\hspace{0.33em}}\mathrm{a}+\mathrm{b}{, find a and b. }\end{array}$

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Polynomials

Medium

Q.

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

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Polynomials

Medium

Q.

$\begin{array}{l}{If the polynomial\hspace{0.33em}}{\mathrm{x}}^{4}-6{\mathrm{x}}^{3}+16{\mathrm{x}}^{2}-25\mathrm{x}+10{\hspace{0.33em}is divided}\\ {by another polynomial\hspace{0.33em}}{\mathrm{x}}^{2}-2\mathrm{x}+\mathrm{k}{,\hspace{0.33em}the remainder comes}\\ {out to be\hspace{0.33em}}\mathrm{x}+\mathrm{a}{,\hspace{0.33em}find\hspace{0.33em}}\mathrm{k}{\hspace{0.33em}and\hspace{0.33em}}\mathrm{a}{.}\end{array}$

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Polynomials

Medium

Q.

Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial $$\rm f(x)=a x^2+b x+c, a \neq 0, c \neq 0 .$$​

[CBSE - 2020]

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Polynomials

Medium

Q.

Divide the polynomial $$\rm f(x)=3 x^2-x^3-3 x+5$$​ by the polynomial $$\rm g(x)=x-1-x^2$$​ and verify the division algorithm.

[CBSE - 2020]

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Polynomials

Medium

Q.

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = x^2 + x - 1$, then determine the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.

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Polynomials

Medium

Q.

If $\alpha$ and $\beta$ are the zeroes of the polynomial $3x^2+6x+k$, and it is given that $\alpha+\beta +\alpha\beta = – \frac{2}{3}$, then find the value of $k$.

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These NCERT Solutions are in line with the latest CBSE syllabus and provide easy-to-understand explanations for polynomial factorization and finding zeroes, making it easier for students to practice and score well in exams.

Q1. Divide the polynomial $p(x)$

$p(x)$ by the polynomial $g(x)$

$g(x)$ and find the quotient and remainder in each of the following:

Solution:

• (i)

  $p(x) = x^3 - 3x^2 + 5x - 3$p(x)=x3−3x2+5x−3,

  $g(x) = x^2 - 2x - 2$g(x)=x2−2x−2

We divide

$p(x)$

p(x) by

$g(x)$

g(x). The quotient is

$x - 3$

x−3 and the remainder is

$7x - 9$

7x−9.

• (ii)

  $p(x) = x^3 - 3x^2 + 5x + 6$p(x)=x3−3x2+5x+6,

  $g(x) = x^2 - 2x - 2$g(x)=x2−2x−2

After division, the quotient is

$x + 3$

x+3 and the remainder is

$8x$

8x.

• (iii)

  $p(x) = 5x^3 - 6x^2 + 7x - 3$p(x)=5x3−6x2+7x−3,

  $g(x) = x^2 - 3x - 2$g(x)=x2−3x−2

The quotient is

$5x - 3$

5x−3, and the remainder is

$5x + 10$

5x+10.

Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

Solution:

• (i)

  $p(x) = x^3 - 2x^2 + 3x + 5$p(x)=x3−2x2+3x+5,

  $g(x) = x^2 - x - 3$g(x)=x2−x−3

When dividing, we get the quotient as

$x - 1$

x−1 and the remainder as

$-5x + 10$

−5x+10.
Since the remainder is not zero,

$g(x)$

g(x) is not a factor of

$p(x)$

p(x).

• (ii)

  $p(x) = x^3 + 5x^2 + 7x + 2$p(x)=x3+5x2+7x+2,

  $g(x) = x^2 + 2x - 1$g(x)=x2+2x−1

After division, the quotient is

$x + 3$

x+3 and the remainder is zero.
Thus,

$g(x)$

g(x) is a factor of

$p(x)$

p(x).

• (iii)

  $p(x) = x^3 + 5x^2 + 3x + 1$p(x)=x3+5x2+3x+1,

  $g(x) = x^2 - x + 1$g(x)=x2−x+1

The quotient is

$x + 3$

x+3, and the remainder is zero.
Hence,

$g(x)$

g(x) is a factor of

$p(x)$

p(x).

Q3. Obtain all other zeroes of $3x^4 + 6x^3 - 2x^2 - 10x - 5$

3x4+6x3−2x2−10x−5 if two of its zeroes are $\frac{5}{3}$

35​ and $-\frac{5}{3}$

−35​.

Solution:

Given that

$\frac{5}{3}$

35​ and

$-\frac{5}{3}$

−35​ are two zeroes of the polynomial,

$x - \frac{5}{3}$

x−35​ and

$x + \frac{5}{3}$

x+35​ are factors of the polynomial.

We divide the given polynomial by

$(x - \frac{5}{3})(x + \frac{5}{3})$

(x−35​)(x+35​), which simplifies to

$x^2 - \left(\frac{5}{3}\right)^2 = x^2 - \frac{25}{9}$

x2−(35​)2=x2−925​.

After dividing, we get the other zeroes of the polynomial as

$1$

1 and

$-1$

−1. Therefore, the zeroes of the given polynomial are

$\frac{5}{3}, -\frac{5}{3}, 1, -1$

35​,−35​,1,−1.

Q4. On dividing $x^3 - 3x^2 + x + 2$

x3−3x2+x+2 by a polynomial $g(x)$

$g(x)$, the quotient and remainder were $x - 2$

$x-2$ and $-2x + 4$

$-2x+4$, respectively. Find $g(x)$

$g(x)$.

Solution:

We use the division algorithm to solve this problem:

$$\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$$

Dividend=Divisor×Quotient+Remainder

Given:

• Dividend

  $= x^3 - 3x^2 + x + 2$=x3−3x2+x+2

• Quotient

  $= x - 2$=x−2

• Remainder

  $= -2x + 4$=−2x+4

Substitute these values into the division algorithm:

$$x^3 - 3x^2 + x + 2 = g(x)(x - 2) + (-2x + 4)$$

x3−3x2+x+2=g(x)(x−2)+(−2x+4)

Simplifying the equation, we find

$g(x) = x^2 - x + 1$

g(x)=x2−x+1.

Q5. Give examples of polynomials

$p(x)$

$p(x)$,

$g(x)$

$g(x)$,

$q(x)$

$q(x)$, and

$r(x)$

$r(x)$, which satisfy the division algorithm and:

• (i)

  $\deg(p(x)) = \deg(q(x))$deg(p(x))=deg(q(x))

• (ii)

  $\deg(q(x)) = \deg(r(x))$deg(q(x))=deg(r(x))

• (iii)

  $\deg(r(x)) = 0$deg(r(x))=0

Solution:

• (i)

  $p(x) = 6x^2 + 2x + 4$p(x)=6x2+2x+4,

  $g(x) = 2$g(x)=2,

  $q(x) = 3x^2 + x + 2$q(x)=3x2+x+2,

  $r(x) = 0$r(x)=0
  The division algorithm is satisfied as the degree of

  $p(x)$p(x) is the same as that of

  $q(x)$q(x).

• (ii)

  $p(x) = x^3 + 2x^2 + 3x + 4$p(x)=x3+2x2+3x+4,

  $g(x) = x^2 + 1$g(x)=x2+1,

  $q(x) = x + 3$q(x)=x+3,

  $r(x) = x + 1$r(x)=x+1
  Here, the degree of

  $q(x)$q(x) is the same as that of

  $r(x)$r(x), and the division algorithm holds.

• (iii)

  $p(x) = x^3 + 3x^2 + 2x + 1$p(x)=x3+3x2+2x+1,

  $g(x) = x$g(x)=x,

  $q(x) = x^2 + 3x + 2$q(x)=x2+3x+2,

  $r(x) = 0$r(x)=0
  The degree of

  $r(x)$r(x) is 0 (since it's a constant), and the division algorithm is satisfied.

FAQs: Class 10 Maths Chapter 2 – Polynomials Exercise 2.3

Q1. How do I find the zeroes of a quadratic polynomial?
Answer:
To find the zeroes of a quadratic polynomial, solve the equation ax² + bx + c = 0 using factorization, completing the square, or the quadratic formula. The zeroes are the values of x for which the polynomial equals 0.

Q2. What is the relationship between zeroes and coefficients of a quadratic polynomial?
Answer:
For a quadratic polynomial ax² + bx + c, the sum and product of its zeroes can be related to the coefficients using:

• Sum of zeroes = -b/a

• Product of zeroes = c/a

Q3. How do I factorize a polynomial in this exercise?
Answer:
To factorize a polynomial, express it as a product of its factors. For example, for x² - 5x + 6, factor it as (x - 2)(x - 3). This can be done by splitting the middle term or using trial and error based on the sum and product of the coefficients.

Q4. Are there any tips for solving the problems in Exercise 2.3?
Answer:

• Use the relationship between the coefficients and zeroes for easier calculations.

• Factorize quadratics using methods like splitting the middle term.

• Double-check your factorization by multiplying the factors back to confirm the polynomial.

Q5. How do NCERT Solutions help in exam preparation?
Answer:
These solutions provide step-by-step guidance to find the zeroes and factorize polynomials, which are key aspects in the Class 10 exams. By practicing these solutions, students can improve their problem-solving skills and understand the concepts thoroughly.