Question

find factors or route of (2x2-3x-2)(2x2-3x)-63​

Answer 1

Answer:(2x+3) (x-3) (2x^2-3x+7)

Step-by-step explanation:

(2x2-3x-2)(2x2-3x)-63

Put 2x^2-3x= y, we get,

(y - 2) y -63

= y^2-2y-63

= y^2 -9y+7y-63= y(y-9)+7(y-9)

= (y-9)(y+7) and putting value of y,

= (2x^2-3x-9) (2x^2-3x+7)

=[ 2x^2-6x+3x-9](2x^2-3x+7)

=[2x(x-3) +3(x-3)](2x^2-3x+7)

= (2x+3) (x-3) (2x^2-3x+7)

Answer 2

$\huge\boxed{\blue{Factor \: Route}}$

$let \: (2 {x}^{2} - 3x) = t$

$(t - 2)t - 63$

${t}^{2} - 2t - 63$

${t}^{2} - 9t + 7t - 63$

$t(t - 9) + 7(t - 9)$

$(t + 7)(t - 9)$

$(2 {x}^{2} - 3 + 7)(2 {x}^{2} - 3x - 9)$

$(2 {x}^{2} - 3x + 7)(2 {x}^{2} - 6x + 3x - 9$

$(2 {x}^{2} - 3x + 9) \: (2x(x - 3) + 3(x - 3))$

$(2 {x}^{2} - 3x + 7) \: (2x + 3)(x - 3)$