NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials Exercise 2.4

NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials Exercise 2.4 are designed to help students with Class 10 exam preparations. These solutions are created by subject experts to explain how to find the zeroes of polynomials and understand the factorization process effectively. The solutions in this exercise focus on solving problems related to polynomials of higher degree and finding their zeroes using algebraic methods.

Exercise 2.4 covers problems that require students to find the zeroes of polynomials and relate them to the factors of the given polynomial. These exercises build a strong foundation for solving real-world algebraic problems and are crucial for mastering the concepts of polynomials in Class 10.

NCERT Solutions for Class 10 Maths Chapter 2 – Polynomials Exercise 2.4

Polynomials

Medium

Q.

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

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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.

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

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Polynomials

Medium

Q.

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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The solutions are aligned with the latest CBSE syllabus, providing step-by-step explanations and easy-to-understand approaches for solving polynomial problems.

Q1. Verify that the numbers given alongside the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

• (i)

  $2x^3 + x^2 - 5x + 2; \quad \frac{1}{2}, 1, -2$2x3+x2−5x+2;21​,1,−2

• (ii)

  $x^3 - 4x^2 + 5x - 2; \quad 2, 1, 1$x3−4x2+5x−2;2,1,1

Solution:

For each polynomial, substitute the given zeroes into the polynomial equation. If they satisfy the equation (i.e., the result is 0), then the given numbers are indeed the zeroes.

For (i):
The polynomial is

$2x^3 + x^2 - 5x + 2$

2x3+x2−5x+2.
For zeroes

$\frac{1}{2}, 1, -2$

21​,1,−2:

1. Put

   $x = \frac{1}{2}$x=21​:

   $p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 - 5\left(\frac{1}{2}\right) + 2 = 0$p(21​)=2(21​)3+(21​)2−5(21​)+2=0
   Therefore,

   $\frac{1}{2}$21​ is a zero of the polynomial.

2. Put

   $x = 1$x=1:

   $p(1) = 2(1)^3 + (1)^2 - 5(1) + 2 = 0$p(1)=2(1)3+(1)2−5(1)+2=0
   Therefore,

   $1$1 is a zero of the polynomial.

3. Put

   $x = -2$x=−2:

   $p(-2) = 2(-2)^3 + (-2)^2 - 5(-2) + 2 = 0$p(−2)=2(−2)3+(−2)2−5(−2)+2=0
   Therefore,

   $-2$−2 is a zero of the polynomial.

Q2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.

Solution:

We know the general equation of a cubic polynomial:

$$p(x) = ax^3 + bx^2 + cx + d$$

p(x)=ax3+bx2+cx+d

The relations between the coefficients and zeroes are:

• Sum of zeroes

  $\alpha + \beta + \gamma = -\frac{b}{a}$α+β+γ=−ab​

• Sum of product of zeroes taken two at a time

  $\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$αβ+βγ+γα=ac​

• Product of zeroes

  $\alpha\beta\gamma = -\frac{d}{a}$αβγ=−ad​

Given:

• Sum of zeroes

  $\alpha + \beta + \gamma = 2$α+β+γ=2

• Sum of product of zeroes taken two at a time

  $\alpha\beta + \beta\gamma + \gamma\alpha = -7$αβ+βγ+γα=−7

• Product of zeroes

  $\alpha\beta\gamma = -14$αβγ=−14

Let

$a = 1$

a=1 (for simplicity), we get:

• $b = -2$b=−2

• $c = -7$c=−7

• $d = 14$d=14

Thus, the cubic polynomial is:

$$p(x) = x^3 - 2x^2 - 7x + 14$$

p(x)=x3−2x2−7x+14

Q3. If the zeroes of the polynomial $x^3 - 3x^2 + x + 1$

x3−3x2+x+1 are $a-b$

$a-b$, $a$

$a$, and $a+b$

$a+b$, find $a$

$a$ and $b$

$b$.

Solution:

The given polynomial is

$p(x) = x^3 - 3x^2 + x + 1$

p(x)=x3−3x2+x+1.

• Sum of zeroes

  $\alpha + \beta + \gamma = -\frac{q}{a}$α+β+γ=−aq​

• Given the zeroes are

  $a-b$a−b,

  $a$a, and

  $a+b$a+b, the sum of the zeroes is:

   

  $$(a-b) + a + (a+b) = 3a$$(a−b)+a+(a+b)=3a

  From the polynomial, we know the sum of zeroes is 3. Hence:

   

  $$3a = 3 \quad \Rightarrow \quad a = 1$$3a=3⇒a=1

• The product of the zeroes is given by:

   

  $$(a-b)(a)(a+b) = a(a^2 - b^2) = a^3 - ab^2$$(a−b)(a)(a+b)=a(a2−b2)=a3−ab2

  The constant term is 1, so:

   

  $$a^3 - ab^2 = 1$$a3−ab2=1

  Substituting

  $a = 1$a=1, we get:

   

  $$1 - b^2 = 1 \quad \Rightarrow \quad b^2 = 0 \quad \Rightarrow \quad b = 0$$1−b2=1⇒b2=0⇒b=0

Thus,

$a = 1$

a=1 and

$b = 0$

b=0.

Q4. If two zeroes of the polynomial $x^4 - 6x^3 - 26x^2 + 138x - 35$

x4−6x3−26x2+138x−35 are $2 \pm \sqrt{3}$

2±3​, find the other zeroes.

Solution:

The given polynomial is

$p(x) = x^4 - 6x^3 - 26x^2 + 138x - 35$

p(x)=x4−6x3−26x2+138x−35.

Since two zeroes are

$2 + \sqrt{3}$

2+3​ and

$2 - \sqrt{3}$

2−3​, the factor corresponding to these zeroes is:

$$(x - (2 + \sqrt{3}))(x - (2 - \sqrt{3})) = (x - 2)^2 - (\sqrt{3})^2 = (x - 2)^2 - 3 = x^2 - 4x + 1$$

(x−(2+3​))(x−(2−3​))=(x−2)2−(3​)2=(x−2)2−3=x2−4x+1

Now divide the given polynomial by

$x^2 - 4x + 1$

x2−4x+1 to find the other factors. After performing the division, we find that the other zeroes are

$x = 7$

x=7 and

$x = -5$

x=−5.

Thus, the zeroes of the polynomial are

$2 + \sqrt{3}, 2 - \sqrt{3}, 7, -5$

2+3​,2−3​,7,−5.

Q5. If the polynomial $x^4 - 6x^3 + 16x^2 - 25x + 10$

x4−6x3+16x2−25x+10 is divided by another polynomial $x^2 - 2x + k$

x2−2x+k, the remainder comes out to be $x + a$

$x+a$, find $k$

$k$ and $a$

$a$.

Solution:

We use the division algorithm for this. The polynomial

$p(x)$

p(x) is divided by

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

g(x)=x2−2x+k with the remainder

$r(x) = x + a$

r(x)=x+a.

By the division algorithm:

$$p(x) = g(x) \times q(x) + r(x)$$

p(x)=g(x)×q(x)+r(x)

Substitute the known values and simplify to solve for

$k$

k and

$a$

a. After solving the equation, we get:

$$k = 5 \quad \text{and} \quad a = -5$$

k=5anda=−5

FAQs: Class 10 Maths Chapter 2 – Polynomials Exercise 2.4

Q1. How do I find the zeroes of a polynomial in Exercise 2.4?
Answer:
To find the zeroes of a polynomial, solve the equation p(x) = 0 by factorizing the polynomial, or use methods such as splitting the middle term or the quadratic formula (for quadratic polynomials). The values of x that satisfy the equation are the zeroes.

Q2. What is the importance of factorizing a polynomial?
Answer:
Factorizing a polynomial helps in finding its zeroes, and it also simplifies the process of solving polynomial equations. Factorization is key to understanding the roots of polynomials, which are essential in algebra.

Q3. What is the relationship between the zeroes and coefficients of a polynomial?
Answer:
For a quadratic polynomial ax² + bx + c, the sum and product of the zeroes are related to the coefficients as follows:

• Sum of zeroes = -b/a

• Product of zeroes = c/a

Q4. How do I solve polynomial equations involving higher degrees?
Answer:
For polynomials of higher degree, try to factorize the polynomial or use methods like synthetic division or the Rational Root Theorem to find the zeroes.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide clear, step-by-step explanations that help students understand how to find the zeroes of polynomials and factorize polynomials effectively. By practicing these solutions, students can improve their problem-solving skills and prepare well for board exams.