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Class 8 RD Sharma Solutions – Chapter 8 Division Of Algebraic Expressions – Exercise 8.6

  • Last Updated : 01 Dec, 2020

Question 1: Divide x2 – 5x + 6 by (x – 3)

Solution:

(x2 – 5x + 6)/(x-3)

Factorise the numerator and then divide it by (x-3):

x2 – 5x + 6

= x2 – 3x – 2x + 6



= (x2 – 3x – 2x + 6)/(x – 3)

= (x(x – 3) – 2(x – 3))/(x – 3)

= ((x – 3)(x – 2))/(x – 3)

= (x – 2)

Therefore, the answer is (x-2).

Question 2: Divide ax2 – ay2 by (ax + ay)

Solution:

(ax2 – ay2)/(ax + ay)

= a(x2 – y2)/(ax + ay)



= a(x – y)(x + y)/a(x + y)

= x – y

Therefore, the answer is (x – y).

Question 3: Divide (x4 – y4) by (x2 – y2)

Solution:

(x4 – y4)/(x2 – y2)

= ((x2)2 – (y2)2)/(x2 – y2)

= ((x2 – y2) (x2 + y2)) / (x2 – y2)

= x2 + y2

Therefore, the answer is (x2 + y2).

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

Solution:



(acx2 + (bc + ad)x + bd) / (ax + b)

= (acx2 + bcx + adx + bd) / (ax + b)

= (cx(ax + b) + d(ax + b)) / (ax + b)

= (ax + b)(cx + d) / (ax + b)

= cx + d

Therefore, the answer is (cx + d).

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

Solution:

((a2 + 2ab + b2) – (a2 + 2ac + c2)) / (2a + b + c)

= ((a + b)2 – (a + c)2) / (2a + b + c)

= ((a + b + a + c)(a + b – a – c)) / (2a + b + c)

= (2a + b + c)(b – c) / (2a + b + c)

= b – c

Therefore, the answer is (b – c).

Question 6: Divide ((1/4)x2 – (1/2)x – 12) by ((1/2)x – 4)

Solution:

(1/4)x2 – (1/2)x – 12

= (1/4)(x2 – 2x – 48)

Now factorize it:

= (1/4)(x2 – 8x + 6x – 48)

= (1/4)(x(x – 8) + 6(x – 8))

= (1/4)(x – 8)(x + 6)



Now divide it by (1/2)x – 4:

= (1/4)(x – 8)(x + 6) / ((1/2)x – 4)

= (1/4)(x – 8)(x + 6) / (1/2)(x – 8)

= (1/4)(2/1)(x + 6)

= (1/2)x + 3

Therefore, the answer is (1/2)x + 3.

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.

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