Question

Factorise:12(x^3+7)^2-8(x^2+7)(2x-1)-15(2x-1)^2

Answer 1

Answer:

Factorise 12(x2+7x)2−8(x2+7x)(2x−1)−15(2x−1)2

Taking a=x2+7x and b=2x−1 , we have

12a2−8ab−15b2

Then, we can proceed to factorise by middle-term splitting method

The product of first and third term is (12a2)(−15b2)=−180(ab)2

By observation, we can split the middle term −8ab into −18ab and 10ab whose product is the same as that of first and third term

Step-by-step explanation:

How can I factorise 12 (x^2+7x) ^2 -8 (x^2+7x) (2x - 1) -15 (2x -1) ^2?

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Factorise 12(x2+7x)2−8(x2+7x)(2x−1)−15(2x−1)2

Taking a=x2+7x and b=2x−1 , we have

12a2−8ab−15b2

Then, we can proceed to factorise by middle-term splitting method

The product of first and third term is (12a2)(−15b2)=−180(ab)2

By observation, we can split the middle term −8ab into −18ab and 10ab whose product is the same as that of first and third term

∴12a2−8ab−15b2

=12a2−18ab+10ab−15b2

=6a(2a−3b)+5b(2a−3b)

=(2a−3b)(6a+5b)

Substituting a=x2+7x and b=2x−1 , we have

(2a−3b)(6a+5b)

=[2(x2+7x)−3(2x−1)][6(x2+7x)+5(2x−1)]

=(2x2+14x−6x+3)(6x2+42x+10x−5)

=(2x2+8x+3)(6x2+52x−5)

∴12(x2+7x)2−8(x2+7x)(2x−1)−15(2x−1)2=(2x2+8x+3)(6x2+52x−5)