RD Sharma Solutions Class 8 Division Of Algebraic Expressions Exercise 8.6

RD Sharma Solutions Class 8 Chapter 8 Exercise 8.6

RD Sharma Class 8 Solutions Chapter 8 Ex 8.6 PDF Free Download

 Question 1: x– 5x + 6 by (x – 3)

 

Soln:

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

 

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

 

= \(\frac{x(x-3)-2(x-2)}{x-3}\)

 

= \(\frac{(x-3)(x-2)}{x-3}\) = x-2

 

Question 2: ax2-ay2  by (ax+ay)

 

Soln:

 

\(\frac{ax^{2}-ay^{2}}{ax+ay}\)

 

= \(\frac{a(x^{2}-y^{2})}{ax+ay}\)

 

= \(\frac{a(x+y)(x-y)}{a(x+y)}\) = x-y

 

 

Question 3: x4-y4 by x2-y2

 

Soln:

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

 

= \(\frac{(x^{2})^{2}-(y^{2})^{2}}{(x^{2}-y^{2})}\)

 

= \(\frac{(x^{2}-y^{2})\times (x^{2}+y^{2})}{(x^{2}-y^{2})}\) = x2+y2

 

 

Question 4: acx2+(bc+ad)x+bd by (ax+b)

 

Soln:

\(\frac{acx^{2}+(bc+ad)x+bd}{ax+b}\)

 

= \(\frac{acx^{2}+bcx+adx+bd}{ax+b}\)

 

= \(\frac{cx(ax+b)+d(ax+b)}{ax+b}\)

 

\(\frac{(ax+b)(cx+d)}{ax+b}\) = cx+d

 

Question 5: (a2+2ab+b2)-(a2+2ac+c2) by (2a+b+c)

 

Soln:

 

\(\frac{(a^{2}+2ab+b^{2})-(a^{2}+2ac+c^{2})}{2a+b+c}\)

 

= \(\frac{(a+b)^{2}-(a+c)^{2}}{2a+b+c}\)

 

= \(\frac{(a+b+a+c)(a+b-a-c}{2a+b+c}\)

 

= \(\frac{(2a+b+c)(b-c)}{2a+b+c}\) = b-c

 

Question 6: \((\frac{1}{4}x^{2}-\frac{1}{2}x-12)\,by\,(\frac{1}{2}x-4)\)

 

Soln:

 

\(\frac{\frac{1}{4}x^{2}-\frac{1}{2}x-12}{\frac{1}{2}x-4}\)

 

= \(\frac{\frac{1}{2}x(\frac{1}{2}x-4)+3()}{\frac{1}{2}x-4}\)

 

= \(\frac{(\frac{1}{2}x+3)(\frac{1}{2}x-4)}{\frac{1}{2}x-4}\) = \((\frac{1}{2}x+3)\)

 

 


Practise This Question

In a marathon of 5 km, Manish could run for 2 km.If  Kushagra ran for 12 of the total length of the marathon, who covered more distance?