Class 8 RD Sharma Solutions – Chapter 8 Division Of Algebraic Expressions – Exercise 8.5

Last Updated : 11 Feb, 2021

Question 1: Divide the first polynomial by the second polynomial in each of the following. Also, write the quotient and remainder:

(i) 3x² + 4x + 5, x – 2

Solution:

3x² + 4x + 5, x – 2
By using factorization method,
$⇒\:\frac{\left(3x^2+4x+5\right)}{x-2} = \:\frac{3x\left(x-2)+10(x-2\right)+25}{x-2}$
$=\:\frac{\left(x-2)(3x+10\right)+25}{x-2}$ (Taking common (x-2) factor)
$=(3x+10)+\:\frac{25}{x-2}$
∴ the Quotient is 3x + 10 and the Remainder is 25.

(ii) 10x² – 7x + 8, 5x – 3

Solution:

10x² – 7x + 8, 5x – 3
By using factorization method,
$⇒\:\frac{\left(10x^2-7x+8\right)}{5x-3} = \:\frac{\left(2x(5x-3)-\:\frac{1}{5}(5x-3)+\:\frac{37}{5}\right)}{5x-3}$
$=\:\frac{\left((5x-3)(2x-\:\frac{1}{5})+\:\frac{37}{5}\right)}{5x-3}$ (Taking common (5x-3) factor)
$=(2x-\:\frac{1}{5})+(\:\frac{\:\frac{37}{5}}{5x-3})$
∴ the Quotient is (2x – 1/5) and the Remainder is 37/5.

(iii) 5y³ – 6y² + 6y – 1, 5y – 1

Solution:

5y³ – 6y² + 6y – 1, 5y – 1
By using factorization method,
$⇒\:\frac{\left(5y^3-6y^2+6y-1\right)}{5y-1}= \:\frac{y^2\left(5y-1)-y(5y-1)+1(5y-1\right)}{5y-1}$
$=\:\frac{(5y-1)(y^2-y+1)}{5y-1}$ (Taking common (5y-1) factor)
$=(y^2-y+1)$
∴ the Quotient is (y² – y + 1) and the Remainder is 0.

(iv) x⁴ – x³ + 5x, x – 1

Solution:

x⁴ – x³ + 5x, x – 1
By using factorization method,
$⇒\:\frac{\left(x^4-x^3+5x\right)}{x-1} = \:\frac{x^3\left(x-1)+5(x-1\right)+5}{x-1}$
$=\:\frac{\left(x-1\right)(x^3+5)+5}{x-1}$ (Taking common (x-1) factor)
$=(x^3+5)+\:\frac{5}{x-1}$
∴ the Quotient is x³ + 5 and the Remainder is 5.

(v) y⁴ + y², y² – 2

Solution:

y⁴ + y², y² – 2
By using factorization method,
$⇒\:\frac{\left(y^4+y^2\right)}{y^2-2}=\:\frac{y^2(y^2-2)+3(y^2-2)+6}{y^2-2}$
$=\:\frac{(y^2-2)(y^2+3)+6}{y^2-2}$ (Taking common (y²-2) factor)
$=(y^2+3)+\:\frac{6}{y^2-2}$
∴ the Quotient is y² + 3 and the Remainder is 6.

Question 2: Find whether or not the first polynomial is a factor of the second:

(i) x + 1, 2x² + 5x + 4

Solution:

x + 1, 2x² + 5x + 4
Let us perform factorization method,
$⇒\:\frac{\left(2x^2+5x+4\right)}{x+1}= \:\frac{2x\left(x+1)+3(x+1\right)+1}{x+1}$
$=\:\frac{(x+1)(2x+3)+1}{x+1}$ (Taking common (x+1) factor)
$=(2x+3)+\:\frac{1}{x+1}$
Since remainder is 1, therefore the first polynomial is not a factor of the second polynomial.

(ii) y – 2, 3y³ + 5y² + 5y + 2

Solution:

y – 2, 3y³ + 5y² + 5y + 2
Let us perform factorization method,
$⇒\:\frac{(3y^3+5y^2+5y+2)}{y-2}=\:\frac{3y^2(y-2)+11y(y-2)+27(y-2)+56}{y-2}$
$=\:\frac{\left((y-2)(3y^2+11y+27)+56\right)}{y-2}$ (Taking common (y-2) factor)
$=(3y^2+11y+27)+\:\frac{56}{y-2}$
Since remainder is 56 therefore the first polynomial is not a factor of the second polynomial.

(iii) 4x² – 5, 4x⁴ + 7x² + 15

Solution:

4x² – 5, 4x⁴ + 7x² + 15
Let us perform factorization method,
$⇒\:\frac{\left(4x^4+7x^2+15\right)}{4x^2-5}=\:\frac{x^2(4x^2-5)+3(4x^2-5)+30}{4x^2-5}$
$=\:\frac{(4x^2-5)(x^2+3)+30}{4x^2-5}$ (Taking common (4x²-5) factor)
$=(x^2+3)+\:\frac{30}{4x^2-5}$
Since remainder is 30 therefore the first polynomial is not a factor of the second polynomial.

(iv) 4 – z, 3z² – 13z + 4

Solution:

4 – z, 3z² – 13z + 4
Let us perform factorization method,
$⇒\:\frac{\left(3z^2-13z+4\right)}{4-z} = \:\frac{\left(3z^2-12z-z+4\right)}{4-z}$
$=\:\frac{3z\left(z-4)-1(z-4\right)}{4-z}$ (Taking common (z-4) factor)
$=\:\frac{\left(z-4)(3z-1\right)}{4-z}$
$=\:\frac{\left(4-z)(1-3z\right)}{4-z}$
$=1-3z$
Since remainder is 0 therefore the first polynomial is a factor of the second polynomial.

(v) 2a – 3, 10a² – 9a – 5

Solution:

2a – 3, 10a² – 9a – 5
Let us perform factorization method,
$⇒\:\frac{\left(10a^2-9a-5\right)}{2a-3}=\:\frac{5a\left(2a-3)+3(2a-3\right)+4}{2a-3}$
$=\:\frac{\left(2a-3)(5a+3\right)+4}{2a-3}$ (Taking common (2a-3) common)
$=(5a+3)+\:\frac{4}{2a-3}$
Since remainder is 4 therefore the first polynomial is not a factor of the second polynomial.

(vi) 4y + 1, 8y² – 2y + 1

Solution:

4y + 1, 8y² – 2y + 1
Let us perform factorization method,
$⇒\:\frac{\left(8y^2-2y+1\right)}{4y+1}=\:\frac{2y\left(4y+1)-1(4y+1\right)+2}{4y+1}$
$=\:\frac{\left(4y+1)(2y-1\right)+2}{4y+1}$ (Taking common (4y+1) factor)
$=(2y-1)+\:\frac{2}{4y+1}$
Since remainder is 2 therefore the first polynomial is not a factor of the second polynomial.