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

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


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If I sell 45th of 10 apples, how many are left?