RD Sharma Solutions Class 8 Division Of Algebraic Expressions Exercise 8.5

RD Sharma Solutions Class 8 Chapter 8 Exercise 8.5

RD Sharma Class 8 Solutions Chapter 8 Ex 8.5 PDF Free Download

RD Sharma Solutions Class 8 Chapter 8 Exercise 8.5

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

(i) \(\frac{3x^{2}+4x+5}{x-2}\)

Soln:

\(\frac{3x^{2}+4x+5}{x-2}\)

= \(\frac{3x(x-2)+10(x-2)+25}{x-2}\)

= \(\frac{3x(x-2)+10(x-2)+25}{x-2}\)

= \((3x+10)+\frac{25}{x-2}\)

Therefore,

Quotient = 3x+10 and remainder = 25

(ii) \(\frac{10x^{2}-7x+8}{5x-3}\)

Soln:

\(\frac{2x(5x-3)-\frac{1}{5}(5x-3)+\frac{47}{5}}{5x-3}\)

= \(\frac{(5x-3)(2x-\frac{1}{5})(5x-3)+\frac{47}{5}}{5x-3}\)

= \((2x-\frac{1}{5})+\frac{\frac{47}{5}}{5x-3}\)

Therefore , quotient= \((2x-\frac{1}{5})\) and remainder = \(\frac{47}{5}\)

(iii) \(\frac{5y^{3}-6y^{2}+6y-1}{5y-1}\)

Soln:

\(\frac{5y^{3}-6y^{2}+6y-1}{5y-1}\)

= \(\frac{y^{2}(5y-1)-y(5y-1)+1(5y-1)}{5y-1}\)

= \(\frac{(5y-1)(y^{2}-y+1)}{5y-1}\)

= y2-y+1+5

Therefore quotient = y2-y+1

And remainder = 0

(iv) \(\frac{x^{4}-x^{3}+5x}{x-1}\)

Soln:

\(\frac{x^{3}(x-1)+5(x-1)+5}{x-1}\)

= \(\frac{(x^{3}+5)(x-1)+5}{x-1}\)

= \((x^{3}+5)+\frac{5}{x-1}\)

Therefore, quotient = x3+5 and remainder= 5

(v) \(\frac{(y^{4}+y^{2})}{y^{2}-2}\)

Soln:

\(\frac{(y^{4}+y^{2})}{y^{2}-2}\)

= \(\frac{y^{2}(y^{2}-2)+3(y^{2}-2)+6}{y^{2}-2}\)

= \(\frac{(y^{2}-2)(y^{2}+3)}{y^{2}-2}\)

\((y^{2}+3)+\frac{6}{y^{2}-2}\)

Therefore, quotient = y3+3 and remainder = 6

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

(i) \(\frac{2x^{2}+5x+4}{x+1}\)

Soln:

= \(\frac{2x(x+1)+3(x+1)+1)}{x+1}\)

= \(\frac{(x+1)(2x+3)+1)}{x+1}\)

Therefore, (x+1) is not a factor of 2x2+5x+4

(ii) \(\frac{3y^{3}+5y^{2}+5y+2}{y-2}\)

Soln:

\(\frac{3y^{3}+5y^{2}+5y+2}{y-2}\)

= \(\frac{3y^{2}(y-2)+11y(y-2)+27(y-2)+56}{y-2}\)

= \(\frac{(y-2)(3y^{2}+11y+27)+56}{y-2}\)

= (3y2+11y+27) + \(\frac{56}{y-2}\)

Therefore, (y-2) is not a factor of 3y3+5y2+5y+2

(iii) \(\frac{4x^{4}+12x^{2}+15}{4x^{2}-5}\)

Soln:

\(\frac{4x^{4}+12x^{2}+15}{4x^{2}-5}\)

= \(\frac{x^{2}(4x^{2}-5)+3(4x^{2}-5)+30}{4x^{2}-5}\)

= (x2+3) + \(\frac{30}{4x^{2}-5}\)

Therefore , (4x2-5) is not a factor of 4x4+7x2+15

(iv) \(\frac{3z^{2}-13z+4}{4-z}\)

Soln:

\(\frac{3z^{2}-13z+4}{4-z}\)

= \(\frac{3z(z-4)-1(z-4)}{4-z}\)

= \(\frac{(z-4)(3z-1)}{4-z}\)

= \(\frac{(4-z)(1-3z)}{4-z}\)

= 1-3z

Therefore, remainder =0

(4-z) is a factor of the factor of 3z2-13z+4

(v) \(\frac{10a^{2}-9a-5}{2a-3}\)

Soln:

\(\frac{10a^{2}-9a-5}{2a-3}\)

= \(\frac{5a(2a-3)+3(2a-3)}{2a-3}\)

= \(\frac{(2a-3)(5a+3)+4}{2a-3}\)

= (5a+3) + \(\frac{4}{2a-3}\)

Therefore, remainder =4

(2a-3) is not a factor of the equation 10a2-9a-5

(vi) \(\frac{8y^{2}-2y+1}{4y+1}\)

Soln:

= \(\frac{2y(4y+1)-1(4y+1)+2}{4y+1}\)

= \(\frac{(4y+1)(2y-1)+2}{4y+1}\)

= 2y-1 + \(\frac{2}{4y+1}\)

Therefore, remainder =2

(4y+1) is not a factor of 8y2-2y+1


Practise This Question

Tina never understood why her uncle Ramses, an astronomer spends most of his time star gazing. To this, Ramses explained her saying that the position of a star seems to be changing with time. Ramses makes four statements in accordance with the practical application of what Tina studied in class. Which of the following is true with respect to stars and their possibilities of positioning?