In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 2 Polynomials Ex 2.3 Pdf, These solutions are solved subject experts from the latest edition books.

## MP Board Class 10th Maths Solutions Chapter 2 Polynomials Ex 2.3

Question 1.

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x^{3} – 3x^{2} + 5x – 3, g(x) = x^{2} – 2

(ii) p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} + 1 – x

(iii) p(x) = x^{4} – 5x + 6, g(x) = 2 – x^{2}

Solution:

(i) Here, dividend p(x) = x^{3} – 3x^{2} + 5x – 3, and divisor g(x) = x^{2} – 2

∴ We have

Thus, the quotient = (x – 3) and remainder = (7x – 9)

(ii) Here, dividend p(x) = x^{4} – 3x^{2} + 4x + 5 and divisor g(x) = x^{2} + 1 – x = x^{2} – x + 1

∴ We have

Thus, the quotient = (x^{2} + x – 3) and remainder = 8

(iii) Here, dividend, p(x) = x^{4} – 5x + 6 and divisor, g(x) = 2 – x^{2} = – x^{2} + 2

∴ We have

Thus, the quotient = -x^{2} – 2 and remainder = -5x +10

Question 2.

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

(i) t^{2} – 3; 2t^{4} + 3t^{3} -2t^{2} – 9t – 12

(ii) x^{2} + 3x + 1; 3x^{4} + 5x^{3} – 7x^{2} + 2x +2

(iii) x^{3} – 3x + 1; x^{5} – 4x^{3} + x^{2} + 3x + 1

Solution:

(i) Dividing 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12 by t^{2} – 3, we have

∵ Remainder = 0

∴ (t^{2} – 3) is a factor of 2t^{4} + 3t^{3} – 2t^{2} – 9t – 12.

(ii) Dividing 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2 by x^{2} + 3x + 1, we have

∵ Remainder = 0

∴ x^{2} + 3x + 1 is a factor of 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2.

(iii) Dividing x^{5} – 4x^{3} + x^{2} + 3x + 1 by x^{3} – 3x + 1, we get

∵ Remainder = 2, i.e., remainder = 0

∴ x^{3} – 3x + 1 is not a factor of x^{5} – 4x^{3} + x^{2} + 3x +1.

Question 3.

Obtain all other zeroes of 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\)

Solution:

We have p(x) = 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5.

Given \(\sqrt{\frac{5}{3}}\) and \(-\sqrt{\frac{5}{3}}\) are zeroes of p(x).

Thus, the other zeroes of the given polynomial are -1 and -1.

Question 4.

On dividing x^{3} – 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).

Solution:

Here, dividend, p(x) = x^{3} – 3x^{2} + x + 2, divisor = g(x), quotient = (x – 2) and remainder = (-2x + 4)

Since, (Quotient × Divisor) + Remainder = Dividend

∴ [(x – 2) × g(x)] + [(-2x + 4)] = x^{3} – 3x^{2} + x + 2

⇒ (x – 2) × g(x)

= x^{3} – 3x^{2} + x + 2 – (-2x + 4)

= x^{3} – 3x^{2} + x + 2 + 2x – 4

= x^{3} – 3x^{2} + 3x – 2

= x^{3} – 3x^{2} + 3x – 2

Thus, the required divisor g(x) = x^{2} – x + 1

Question 5.

Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0

Solution:

(i) p(x) = 3x^{2} – 6x + 27,

g(x) = 3 and q(x) = x^{2} – 2x + 9.

Now, deg p(x) = deg q(x)

r(x) = 0

⇒ p(x) = q(x) × g(x) + r(x)

(ii) p(x) = 2x^{3} – 2x^{2} + 2x + 3,

g(x) = 2x^{2} – 1, and

r(x) = 3x + 2, deg q(x) = deg r(x)

⇒ p(x) = q(x) × g(x) + r(x)

(iii) p(x) = 2x^{3} – 4x^{2} + x + 4,

g(x) = 2x^{2} + 1,

q(x) = x – 2 and r(x) = 6,

deg r(x) = 0

⇒ p(x) = q(x) × g(x) + r(x)