Class 8 RD Sharma Solutions Chapter 8 Division Of Algebraic Expressions  Exercise 8.5 (original) (raw)

Last Updated : 12 Sep, 2024

Introduction

Exercise 8.5 focuses on dividing polynomials by monomials and polynomials. It covers techniques like factoring out common terms, using the distributive property, and long division of polynomials. Students learn to simplify complex fractions and solve problems involving algebraic division.

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

****(i) 3x** 2 + 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) 10x** 2 – 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) 5y** 3 – 6y 2 + 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) x** 4 – x 3 + 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) y** 4 + y 2 , y 2 – 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, 2x** 2 + 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, 3y** 3 + 5y 2 + 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) 4x** 2 – 5, 4x 4 + 7x 2 + 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, 3z** 2 – 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, 10a** 2 – 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, 8y** 2 – 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.

Summary

Summary: Exercise 8.5 builds on students' understanding of polynomial division, teaching them to divide increasingly complex expressions. It reinforces concepts like factoring, the distributive property, and long division of polynomials. This skill is crucial for simplifying algebraic fractions and solving higher-level math problems.