Question

Factorise
12 (x square + 7 x) square minus 8 bracket X square + 7 x bracket close bracket open 2 x minus 1 bracket close -15 bracket open 2x - 1 bracket close square

Answer 1

Answer:

$factor\:12\left(x^2+7x\right)^2-8\left(x+7x\right)+\left(2x-1\right)-15\left(2x-1\right)\\=2\left(6x^4+84x^3+294x^2-46x+7\right)$

Step-by-step explanation:

We can simplify  the expression and we see  further factorization not possible

$12\left(x^2+7x\right)^2-8\left(x+7x\right)+\left(2x-1\right)-15\left(2x-1\right)\\\mathrm{Factor}\:12\left(x^2+7x\right)^2-8\left(x+7x\right):\quad 4x\left(3x\left(x+7\right)^2-16\right)\\\mathrm{Factor}\:\left(2x-1\right)-15\left(2x-1\right):\quad -14\left(-1+2x\right)\\=4x\left(3x\left(x+7\right)^2-16\right)-14\left(-1+2x\right)\\\mathrm{Rewrite\:as}$

$=2\cdot \:2x\left(-16+\left(7+x\right)^2\cdot \:3x\right)-2\cdot \:7\left(-1+2x\right)\\\mathrm{Factor\:out\:common\:term\:}2\\=2\left(2x\left(-16+\left(7+x\right)^2\cdot \:3x\right)-7\left(-1+2x\right)\right)\\\mathrm{Expand}\:2x\left(3x\left(x+7\right)^2-16\right)-7\left(2x-1\right):\quad 6x^4+84x^3+294x^2-46x+7\\=2\left(6x^4+84x^3+294x^2-46x+7\right)$

Answer 2

Answer:

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