In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.4. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 1.

Verify that the numbers given alongside of 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; \(\frac{1}{2}\), 1, -2

(ii) x^{3} – 4x^{2} + 5x – 2; 2, 1, 1

Solution:

Again, p(1) = 2(1)^{3} + (1)^{2} – 5(1) + 2

= 2 + 1 – 5 + 2 = 0

⇒ 1 is a zero of p(x).

Also, p(-2) = 2(-2)^{3} + (-2)^{2} – 5(-2) + 2

= -16 + 4 + 10 + 2 = -16 +16 = 0

= -2 is a zero of p(x).

Now, p(x) = 2x^{3} + x^{2} – 5x + 2

∴ Comparing it with ax^{3} + bx^{2} + cx + d, we have a = 2, b = 1, c = -5, and d = 2

Thus, the relationship between the coefficients and the zeroes of p(x) is verified.

(ii) Here, p(x) = x^{3} – 4x^{2} + 5x – 2

∴ p(2) = (2)^{3} – 4(2)^{2} + 5(2) – 2

= 8 – 16 + 10 – 2 = 18 – 18 = 0

⇒ 2 is a zero of p(x)

Again p(1) = (1)^{3} – 4(1)^{2} + 5(1) – 2

= 1 – 4 + 5 – 2 = 6 – 6 = 0

⇒ 1 is a zero of p(x)

Now, comparing p(x) = x^{3} – 4x^{2} + 5x – 2

with ax^{3} + bx^{2} + cx + d = 0, we have

a = 1, b = -4, c = 5 and d = -2

Also 2, 1 and 1 are the zeroes of p(x).

Let α = 2,

β = 1,

γ = 1

Now, sum of zeroes = α + β + γ

= 2 + 1 + 1 = 4 = -b/a

Thus, the relationship between the zeroes and the coefficients of p(x) is verified.

Question 2.

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:

Let the required cubic polynomial be ax^{3} + 6x^{2} + cx + d = 0 and its zeroes be α, β and γ.

∴ The requied cubic polynomial is

1x^{3} + (-2)x^{2} + (-7)x + 14 = 0

= x^{3} – 2x^{2} – 7x + 14 = 0

Question 3.

If the zeroes of the polynomial x^{3} – 3x^{2} + x + 1 are a -b, a, a + b, find a and b.

Solution:

We have p(x) = x^{3} – 3x^{2} + x + 1

Comparing it with Ax^{3} + Bx^{2} + Cx + D,

We have A = 1, B = -3, C = 1 and D = 1

∵ It is given (a – b), a and (a + b) are the zeroes of the polynomial.

Question 4.

If two zeroes of the polynomial

x^{4} – 6x^{3} – 26x^{2}+ 138x – 35 are 2±\(\sqrt{3}\), find other zeroes.

Solution:

Here, p(x) = x^{4} – 6x^{3} – 26x^{3} + 138x – 35

∵ Two of the zeroes of p(x) are : 2 ± \(\sqrt{3}\)

(x^{2} – 4x + 1)(x^{2} – 2x – 35) = p(x)

⇒ (x^{2} – 4x + 1) (x^{2} – 7x + 5x – 35) = p(x)

⇒ (x^{2} – 4x + 1) [x(x – 7) + 5(x – 7)] = p(x)

⇒ (x^{2} – 4x + 1)(x – 7)(x + 5) = p(x)

i.e., (x – 7) and (x + 5) are other factors of p(x).

∴ 7 and – 5 are other zeroes of the given polynomial.

Question 5.

If the polynomial x^{4} – 6x^{3} + 16x^{2} – 25x + 10 is divided by another polynomial x^{2} – 2x + k, the remainder comes out to be x + a, find k and a.

Solution:

Applying the division algorithm to the polynomials x^{4} – 6x^{3} + 16x^{2} – 25x + 10 and x^{2} – 2x + k, we have

∴ Remainder = (2k – 9)x – k(8 – k) + 10

But the remainder = x + a (Given)

Therefore, comparing them, we have

and a = -k(8 – k) + 10

= -5(8 – 5) + 10

= -5(3) + 10 = -15 + 10 = -5

Thus k = 5 and a = -5

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.4. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 1.

Verify that the numbers given alongside of 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; \(\frac{1}{2}\), 1, -2

(ii) x^{3} – 4x^{2} + 5x – 2; 2, 1, 1

Solution:

Again, p(1) = 2(1)^{3} + (1)^{2} – 5(1) + 2

= 2 + 1 – 5 + 2 = 0

⇒ 1 is a zero of p(x).

Also, p(-2) = 2(-2)^{3} + (-2)^{2} – 5(-2) + 2

= -16 + 4 + 10 + 2 = -16 +16 = 0

= -2 is a zero of p(x).

Now, p(x) = 2x^{3} + x^{2} – 5x + 2

∴ Comparing it with ax^{3} + bx^{2} + cx + d, we have a = 2, b = 1, c = -5, and d = 2

Thus, the relationship between the coefficients and the zeroes of p(x) is verified.

(ii) Here, p(x) = x^{3} – 4x^{2} + 5x – 2

∴ p(2) = (2)^{3} – 4(2)^{2} + 5(2) – 2

= 8 – 16 + 10 – 2 = 18 – 18 = 0

⇒ 2 is a zero of p(x)

Again p(1) = (1)^{3} – 4(1)^{2} + 5(1) – 2

= 1 – 4 + 5 – 2 = 6 – 6 = 0

⇒ 1 is a zero of p(x)

Now, comparing p(x) = x^{3} – 4x^{2} + 5x – 2

with ax^{3} + bx^{2} + cx + d = 0, we have

a = 1, b = -4, c = 5 and d = -2

Also 2, 1 and 1 are the zeroes of p(x).

Let α = 2,

β = 1,

γ = 1

Now, sum of zeroes = α + β + γ

= 2 + 1 + 1 = 4 = -b/a

Thus, the relationship between the zeroes and the coefficients of p(x) is verified.

Question 2.

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:

Let the required cubic polynomial be ax^{3} + 6x^{2} + cx + d = 0 and its zeroes be α, β and γ.

∴ The requied cubic polynomial is

1x^{3} + (-2)x^{2} + (-7)x + 14 = 0

= x^{3} – 2x^{2} – 7x + 14 = 0

Question 3.

If the zeroes of the polynomial x^{3} – 3x^{2} + x + 1 are a -b, a, a + b, find a and b.

Solution:

We have p(x) = x^{3} – 3x^{2} + x + 1

Comparing it with Ax^{3} + Bx^{2} + Cx + D,

We have A = 1, B = -3, C = 1 and D = 1

∵ It is given (a – b), a and (a + b) are the zeroes of the polynomial.

Question 4.

If two zeroes of the polynomial

x^{4} – 6x^{3} – 26x^{2}+ 138x – 35 are 2±\(\sqrt{3}\), find other zeroes.

Solution:

Here, p(x) = x^{4} – 6x^{3} – 26x^{3} + 138x – 35

∵ Two of the zeroes of p(x) are : 2 ± \(\sqrt{3}\)

(x^{2} – 4x + 1)(x^{2} – 2x – 35) = p(x)

⇒ (x^{2} – 4x + 1) (x^{2} – 7x + 5x – 35) = p(x)

⇒ (x^{2} – 4x + 1) [x(x – 7) + 5(x – 7)] = p(x)

⇒ (x^{2} – 4x + 1)(x – 7)(x + 5) = p(x)

i.e., (x – 7) and (x + 5) are other factors of p(x).

∴ 7 and – 5 are other zeroes of the given polynomial.

Question 5.

If the polynomial x^{4} – 6x^{3} + 16x^{2} – 25x + 10 is divided by another polynomial x^{2} – 2x + k, the remainder comes out to be x + a, find k and a.

Solution:

Applying the division algorithm to the polynomials x^{4} – 6x^{3} + 16x^{2} – 25x + 10 and x^{2} – 2x + k, we have

∴ Remainder = (2k – 9)x – k(8 – k) + 10

But the remainder = x + a (Given)

Therefore, comparing them, we have

and a = -k(8 – k) + 10

= -5(8 – 5) + 10

= -5(3) + 10 = -15 + 10 = -5

Thus k = 5 and a = -5