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 #