What is the standard form of $y= (x^2-4)(2x-4) -(2x+5)^2$ ?

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Answer 1 · 2017-12-27T16:05:35

$y=2x^3-8x^2-28x-9$

$y=(x^2-4)(2x-4)-(2x+5)^2$

$y=((x^2)(2x)+(-4)(2x)+(x^2)(-4)+(-4)(-4))-((2x)^2+2(2x)(5)+(5)^2)$

$y=2x^3-8x-4x^2+16-4x^2-20x-25$

$y=2x^3-8x^2-28x-9$

Answer 2 · 2017-12-27T16:06:52

$y=2x^3-8x^2-28x-9$

We have to expand a bunch:

$(x^2-4)(2x-4)=2x^3-4x^2-8x+16$
$(2x+5)^2=4x^2+20x+25$

Start with:
$(x^2-4)(2x-4) - (2x+5)^2$

Replace with expansions:
$= 2x^3-4x^2-8x+16 - (4x^2+20x+25)$

Distribute the negative:
$=2x^3-4x^2-8x+16 - 4x^2-20x-25$

Collect like terms:
$=2x^3-8x^2-28x-9$

So our final answer is $y=2x^3-8x^2-28x-9$

What is the standard form of y= (x^2-4)(2x-4) -(2x+5)^2y= (x^2-4)(2x-4) -(2x+5)^2?