What is the standard form of $y= (2x-3)(4x-4)^2$ ?

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Answer 1 · 2018-05-14T09:18:51

$y=32x^3-112x^2+128x-48$

$y=(2x-3)(4x-4)(4x-4)$
$y=(8x^2-8x-12x+12)(4x-4)$
$y=(8x^2-20x+12)(4x-4)$
$y=32x^3-80x^2+48x-32x^2+80x-48$
$y=32x^3-112x^2+128x-48$

What is the standard form of y= (2x-3)(4x-4)^2 y= (2x-3)(4x-4)^2 ?