Question

factorise 1-2a-2b-3 (a+b)^2

Answer 1

1-2a -2b -3(a+b)²
= 1-2(a+b)-3(a+b)²
=1- 3(a+b) +(a+b) -3(a+b)²
=1[1-3(a+b)]+(a+b)[1-3(a+b)]
=[1-3(a+b)][1+(a+b)]
=(1-3a-3b)(1+a+b)

Answer 2

Answer :

$[1 - 3 a - 3 b] [1 + a + b] \:$

Solution :

$1-2a-2b-3 (a+b)^2 \\ = 1 - 2(a +b) - 3 ( a +b)^2 \\ = 1 - 3(a+b)+ (a+b) - 3(a+b)^2 </p><p> \\ = 1 [1 - 3 (a + b)] + (a +b) [1 - 3 ( a +b)] \\ </p><p> = [1 - 3 (a + b)] [1 + (a + b)] \\ = [1 - 3 a - 3 b] [1 + a + b]$

Detailed explanation of the above factorization :

In step 1, We have written - 2a - 2b as - 2(a+b)

In step 2, We wrote - 2(a+b) as = - 3(a+b)+(a+b) based on the fact, - 3 +1 = - 2

In step 3, then 1, a+b are taken as common from the first 2, last 2 terms respectively.

Next we simplified to get [1 - 3 a - 3 b] [1 + a + b]