Question

Resolve into factors:, (x-5)(x-7)(x+6)(x+4)-504

Answer 1

Solution:-

given by:-

(x-5)(x-7)(x+6)(x+4)-504

$( {x}^{2} - 12x + 35)( {x}^{2} + 10x + 24) - 504 \\ {x}^{4} + 10 {x}^{3} + 24 {x}^{2} - 12 {x}^{3} - 120 {x}^{2} - 288x + 35 {x}^{2} + 350x + 840 - 504 \\ ( {x}^{4} - 2 {x}^{3} - 61 {x}^{2} + 62x + 336)ans$

Answer 2

Q :Factorise
(x-5)(x-7)(x+6)(x+4)-504 :

Final Answer :
(x-3)(x+2)(x-8)(x+7)


Steps:
1) (x-5)(x+4)(x-7)(x+6)-504

$= > ( {x}^{2} - x - 20)( {x}^{2} - x - 42) \\ - 504$
$put \: {x}^{2} - x = t \\ we \: get \\ (t - 20)(t - 42) - 504 \\ = > {t}^{2} - 62t + 840 - 504 \\ = > {t}^{2} - 62t + 336 \\ = > {t}^{2} - 56t - 6t + 336 \\ = > t(t - 56) - 6(t - 56) \\ = > (t - 6)(t - 56)$

2)Substitute, value of t, we get

$( {x}^{2} - x - 6)( {x}^{2} - x - 56) \\ = > ( {x}^{2} - 3x + 2x - 6)( {x}^{2} - 8x \\ + 7x - 56) \\ = > ( x(x - 3) + 2(x - 3)) \times \\ (x -8)(x - 7) \\ = > ( x - 3)(x + 2)(x - 8)(x + 7)$