What is the standard form of y= (x^2-4)(2x-4) -(2x+5)^2y=(x2−4)(2x−4)−(2x+5)2?

2 Answers
YoshiMan007
Dec 27, 2017

y=2x^3-8x^2-28x-9y=2x3−8x2−28x−9

Explanation:

y=(x^2-4)(2x-4)-(2x+5)^2y=(x2−4)(2x−4)−(2x+5)2

y=((x^2)(2x)+(-4)(2x)+(x^2)(-4)+(-4)(-4))-((2x)^2+2(2x)(5)+(5)^2)y=((x2)(2x)+(−4)(2x)+(x2)(−4)+(−4)(−4))−((2x)2+2(2x)(5)+(5)2)

y=2x^3-8x-4x^2+16-4x^2-20x-25y=2x3−8x−4x2+16−4x2−20x−25

y=2x^3-8x^2-28x-9y=2x3−8x2−28x−9

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turksvids
Dec 27, 2017

y=2x^3-8x^2-28x-9 y=2x3−8x2−28x−9

Explanation:

We have to expand a bunch:

(x^2-4)(2x-4)=2x^3-4x^2-8x+16(x2−4)(2x−4)=2x3−4x2−8x+16
(2x+5)^2=4x^2+20x+25(2x+5)2=4x2+20x+25

Start with:
(x^2-4)(2x-4) - (2x+5)^2(x2−4)(2x−4)−(2x+5)2

Replace with expansions:
= 2x^3-4x^2-8x+16 - (4x^2+20x+25)=2x3−4x2−8x+16−(4x2+20x+25)

Distribute the negative:
=2x^3-4x^2-8x+16 - 4x^2-20x-25=2x3−4x2−8x+16−4x2−20x−25

Collect like terms:
=2x^3-8x^2-28x-9 =2x3−8x2−28x−9

So our final answer is y=2x^3-8x^2-28x-9 y=2x3−8x2−28x−9

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