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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

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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.




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