**Previous Exercise – Polynomials – Exercise 8.4 – Class X**

#### Polynomials – Exercise 8.5

- Find the quotients and remainder using synthetic division.

(i) (x^{3} + x^{2} – 3x + 5) /(x – 1)

(ii) (3x^{3} – 2x^{2} + 7x – 5) /(x + 3)

(iii) (4x^{3} – 16x^{2} – 9x – 36)/(x + 2)

(iv) (6x^{4} – 29x^{3} + 40x^{2} – 12)/(x – 3)

(v) (8x^{4} – 27x^{2} + 6x + 9)/(x + 1)

(vi) (3x^{3} – 4x^{2} – 10x + 6)/(3x – 2)

(vii) (8x^{4} – 27x^{2} + 6x – 5)/(4x + 1)

(viii) (2x^{4} – 7x^{3} – 13x^{2} + 63x – 48)/(2x – 1)

- If the quotient obtained on dividng (x
^{4}+ 10x^{3}+ 35x^{2}+ 50x + 29) by (x + 4) is (x^{3}– ax^{2}+ bx + 6) then find a, b and also the remainder - If the quotient obtained on dividing (8x
^{4}– 2x^{2}+ 6x – 7) by (2x +1) is (4x^{3}+ px^{2}– qx + 3) then find p, q and also find the remainder.

#### Polynomials – Exercise 8.5 – Solutions:

**Find the quotients and remainder using synthetic division.**

**(i) (x ^{3} + x^{2} – 3x + 5) /(x – 1)**

Solution:

Therefore, q(x) = x^{2} + 2x – 1 , r(x) = 4

**(ii) (3x ^{3} – 2x^{2} + 7x – 5) /(x + 3)**

Solution:

Therefore, q(x) = 3x^{2} – 11x + 40 and r(x) = -125

**(iii) (4x ^{3} – 16x^{2} – 9x – 36)/(x + 2)**

Solution:

Therefore, q(x) = 4x^{2} – 24x + 39 and r(x) = -114

**(iv) (6x ^{4} – 29x^{3} + 40x^{2} – 12)/(x – 3)**

Solution:

Therefore, q(x) = 6x^{3} – 11x^{2} + 7x + 21 and r(x) = 51

** (v) (8x ^{4} – 27x^{2} + 6x + 9)/(x + 1)**

Solution:

Therefore, q(x) = 8x^{3} – 8x^{2} – 19x + 25 and r(x) = -16

**(vi) (3x ^{3} – 4x^{2} – 10x + 6)/(3x – 2)**

Solution:

Therefore, q(x) = 3x^{2} – 2x – ^{34}/_{3} and r(x) = –^{14}/_{9}

⇒q(x) = x^{2} – ^{2}/_{3} x – ^{34}/_{9} and r(x) = –^{14}/_{9}

**(vii) (8x ^{4} – 2x^{2} + 6x – 5)/(4x + 1)**

Solution:

Therefore, q(x) = 8x^{3} – 2x^{2} – ^{3}/_{2}x + ^{51}/_{8} and r(x) = –^{211}/_{32}

⇒q(x) = 2x^{3} – ^{1}/_{2}x^{2} – ^{3}/_{8 }x + ^{51}/_{32} and r(x) = –^{211}/_{32}

**(viii) (2x ^{4} – 7x^{3} – 13x^{2} + 63x – 48)/(2x – 1)**

Solution:

Therefore, q(x) = 2x^{3} – 6x^{2} – 16x + 55 and r(x) = –^{41}/_{2}

⇒q(x) = x^{3} – 3x^{2} – _{ }8x + ^{55}/_{2} and r(x) = –^{41}/_{2}

**If the quotient obtained on dividing (x**^{4}+ 10x^{3}+ 35x^{2}+ 50x + 29) by (x + 4) is (x^{3}– ax^{2}+ bx + 6) then find a, b and also the remainder

Solution:

Given, if the quotient obtained on dividing x^{4} + 10x^{3}_{ + }35x^{2} + 50x + 29 by (x + 4) is (x^{3} – ax^{2 }+ bx + 6).

q’(x) = x^{3} – ax^{2 }+ bx + 6

On dividing x^{4} + 10x^{3}_{ + }35x^{2} + 50x + 29 by (x + 4) quotient is x^{3} + 6x^{2} + 11x + 6 and the remainder is 5.

i.e., q(x) = x^{3} + 6x^{2} + 11x + 6 and r(x) = 5

q’(x) = q(x)

x^{3} + 6x^{2} + 11x + 6 = x^{3} – ax^{2} + bx + 6

Coefficient of x^{2} 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.

**If the quotient obtained on dividing (8x**^{4}– 2x^{2}+ 6x – 7) by (2x +1) is (4x^{3}+ px^{2}– qx + 3) then find p, q and also find the remainder.

Solution:

Given, if the quotient obtained on dividing (8x^{4} – 2x^{2} + 6x – 7) by (2x +1) is (4x^{3} + px^{2} – qx + 3).

q’(x) = (4x^{3} + px^{2} – qx + 3)

On dividing (8x^{4} – 2x^{2} + 6x – 7) by (2x +1) quotient is 8x^{3} – 4x^{2} + 6 and the remainder is -10

i.e., q(x) = 8x^{3} – 4x^{2} + 6 = 4x^{3} – 2x^{2} + 6 and r(x) = -10

q’(x) = q(x)

4x^{3} + px^{2} – qx + 3 = 4x^{3} – 2x^{2} + 6

Coefficient of x^{2} 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