Question

Factorise 1 - 2 (a + 2b) - 3 (a + 2b)^2​

Answer 1

Answer:

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]

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]

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Answer 2

Answer:

The answer is

(1-3a-6b)(1+a+2b)

Step-by-step explanation:

Let x=a+2b.

Then,

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

= 1 - 2x - 3x^2

= 1 - 3x + x - 3x^2

= (1 - 3x) + x(1 - 3x)

= (1 - 3x) (1 + x)

= {1 - 3(a+2b)} {(1+(a+2b)}. (substituting the value of x)

= (1 - 3a - 6b) (1 + a + 2b)