Factorise completely the expressions.

  1. (a)

    4(x−y)2−9(x+y)24(x - y)^2 - 9(x + y)^2

  2. (b)

    10ax+15ay−8bx−12by10ax + 15ay - 8bx - 12by

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Worked solution

13(a)

1. Rewrite each squared term as a square of a bracket.

4(x−y)2−9(x+y)2=[2(x−y)]2−[3(x+y)]24(x-y)^2-9(x+y)^2=[2(x-y)]^2-[3(x+y)]^2

2. Apply the difference of two squares formula and simplify the brackets.

Using A2−B2=(A−B)(A+B)A^2-B^2=(A-B)(A+B),
[2(x−y)]2−[3(x+y)]2[2(x-y)]^2-[3(x+y)]^2
=[2(x−y)−3(x+y)][2(x−y)+3(x+y)]=[2(x-y)-3(x+y)][2(x-y)+3(x+y)]
=(2x−2y−3x−3y)(2x−2y+3x+3y)=(2x-2y-3x-3y)(2x-2y+3x+3y)
=(−x−5y)(5x+y)=(-x-5y)(5x+y)
=−(x+5y)(5x+y)=-(x+5y)(5x+y)

Answer

(2x−2y−3x−3y)(2x−2y+3x+3y)=−(x+5y)(x−y)(2x-2y-3x-3y)(2x-2y+3x+3y)=-(x+5y)(x-y)

13(b)

1. Group the terms and factorise each pair.

10ax+15ay−8bx−12by=5a(2x+3y)−4b(2x+3y)10ax+15ay-8bx-12by=5a(2x+3y)-4b(2x+3y)

2. Extract the common binomial factor.

5a(2x+3y)−4b(2x+3y)=(5a−4b)(2x+3y)5a(2x+3y)-4b(2x+3y)=(5a-4b)(2x+3y)

Answer

5a(2x+3y)−4b(2x+3y)=(5a−4b)(2x+3y)5a(2x+3y)-4b(2x+3y)=(5a-4b)(2x+3y)

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