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

1 Answer
Mark D.
May 14, 2018

y=32x^3-112x^2+128x-48y=32x3−112x2+128x−48

Explanation:

y=(2x-3)(4x-4)(4x-4)y=(2x−3)(4x−4)(4x−4)
y=(8x^2-8x-12x+12)(4x-4)y=(8x2−8x−12x+12)(4x−4)
y=(8x^2-20x+12)(4x-4)y=(8x2−20x+12)(4x−4)
y=32x^3-80x^2+48x-32x^2+80x-48y=32x3−80x2+48x−32x2+80x−48
y=32x^3-112x^2+128x-48y=32x3−112x2+128x−48

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