# Polynomials – Exercise 8.5 – Class X

Previous Exercise – Polynomials – Exercise 8.4 – Class X

#### Polynomials – Exercise 8.5

1. Find the quotients and remainder using synthetic division.

(i) (x3 + x2 – 3x + 5) /(x – 1)

(ii)  (3x3 – 2x2 + 7x  – 5) /(x + 3)

(iii) (4x3 – 16x2 – 9x – 36)/(x + 2)

(iv) (6x4 – 29x3 + 40x2  – 12)/(x – 3)

(v) (8x4 – 27x2 +  6x + 9)/(x + 1)

(vi) (3x3 – 4x2 –  10x + 6)/(3x – 2)

(vii) (8x4 – 27x2 + 6x – 5)/(4x + 1)

(viii) (2x4 – 7x3 – 13x2 + 63x – 48)/(2x – 1)

1. If the quotient obtained on dividng (x4 + 10x3 + 35x2 + 50x + 29) by (x + 4) is (x3 – ax2 + bx + 6) then find a, b and also the remainder
2. If the quotient obtained on dividing (8x4 – 2x2 + 6x – 7) by (2x +1) is (4x3 + px2 – qx + 3) then find p, q and also find the remainder.

#### Polynomials – Exercise 8.5 – Solutions:

1. Find the quotients and remainder using synthetic division.

(i) (x3 + x2 – 3x + 5) /(x – 1)

Solution: Therefore, q(x) = x2 + 2x – 1 , r(x) = 4

(ii)  (3x3 – 2x2 + 7x  – 5) /(x + 3)

Solution: Therefore, q(x) = 3x2 – 11x + 40 and r(x) = -125

(iii) (4x3 – 16x2 – 9x – 36)/(x + 2)

Solution: Therefore, q(x) =  4x2 – 24x + 39 and r(x) = -114

(iv) (6x4 – 29x3 + 40x2  – 12)/(x – 3)

Solution: Therefore, q(x) = 6x3 – 11x2 + 7x + 21 and r(x) = 51

(v) (8x4 – 27x2 +  6x + 9)/(x + 1)

Solution: Therefore, q(x) = 8x3 – 8x2 – 19x + 25 and r(x) = -16

(vi) (3x3 – 4x2 –  10x + 6)/(3x – 2)

Solution: Therefore, q(x) = 3x2  – 2x – 34/3 and r(x) = –14/9

⇒q(x) = x22/3 x – 34/9 and r(x) = –14/9

(vii) (8x4 – 2x2 + 6x – 5)/(4x + 1)

Solution: Therefore, q(x) = 8x3  – 2x23/2x + 51/8 and r(x) = –211/32

⇒q(x) = 2x3  – 1/2x23/8 x + 51/32 and r(x) = –211/32

(viii) (2x4 – 7x3 – 13x2 + 63x – 48)/(2x – 1)

Solution: Therefore, q(x) = 2x3  – 6x2 – 16x + 55 and r(x) = –41/2

⇒q(x) = x3  – 3x2 8x + 55/2 and r(x) = –41/2

1. If the quotient obtained on dividing (x4 + 10x3 + 35x2 + 50x + 29) by (x + 4) is (x3 – ax2 + bx + 6) then find a, b and also the remainder

Solution: Given, if the quotient obtained on dividing x4 + 10x3 + 35x2 + 50x + 29 by  (x + 4) is (x3 – ax2 + bx + 6).

q’(x) = x3 – ax2 + bx + 6

On dividing x4 + 10x3 + 35x2 + 50x + 29 by  (x + 4) quotient is x3 + 6x2 + 11x + 6 and the remainder is 5.

i.e., q(x) = x3 + 6x2 + 11x + 6 and r(x) = 5

q’(x) = q(x)

x3 + 6x2 + 11x + 6 = x3 – ax2 + bx + 6

Coefficient of x2 on both the sides, 6 = -a ⇒ a = -6

Coefficient of x on both the sides, 11 = b

Therefore, a = -6 and b = 11.

The remainder r(x) = 5.

1. If the quotient obtained on dividing (8x4 – 2x2 + 6x – 7) by (2x +1) is (4x3 + px2 – qx + 3) then find p, q and also find the remainder.

Solution:

Given, if the quotient obtained on dividing (8x4 – 2x2 + 6x – 7) by (2x +1) is (4x3 + px2 – qx + 3).

q’(x) = (4x3 + px2 – qx + 3) On dividing (8x4 – 2x2 + 6x – 7) by (2x +1) quotient is 8x3 – 4x2 + 6 and the remainder is -10

i.e., q(x) = 8x3 – 4x2 + 6 = 4x3 – 2x2 + 6  and r(x) = -10

q’(x) = q(x)

4x3 + px2 – qx + 3 = 4x3 – 2x2 + 6

Coefficient of x2 on both the sides, p = -2

Coefficient of x on both the sides, -q = 0 ⇒ q = 0

Therefore, p = -2 , q = 0 and r(x) = -10