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

In a colony, the number of rooms in an apartment is equal to the number of apartments. If the total number of rooms in the colony is 2401, what is the number of apartments?