SOLUTION: (x^3+8/x^2-4)times(x^2-4x+4/x^2-2x+4) my math teacher said i got this wrong my answer was X(x-2)/-2x+4 he said you didnt solve the difference of squares correctly??
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-> SOLUTION: (x^3+8/x^2-4)times(x^2-4x+4/x^2-2x+4) my math teacher said i got this wrong my answer was X(x-2)/-2x+4 he said you didnt solve the difference of squares correctly??
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Question 382925: (x^3+8/x^2-4)times(x^2-4x+4/x^2-2x+4) my math teacher said i got this wrong my answer was X(x-2)/-2x+4 he said you didnt solve the difference of squares correctly?? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! (x^3+8/x^2-4)times(x^2-4x+4/x^2-2x+4) my math teacher said i got this wrong my answer was X(x-2)/-2x+4 he said you didnt solve the difference of squares correctly??
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This is a "sum" .Factor
x^3+8
= x^3+2^3
= (x+2)(x^2-x+2)
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This is a "difference of squares":
Factor x^2-4 = (x+2)(x-2)
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Your Problem:
(x^3+8/x^2-4)times(x^2-4x+4/x^2-2x+4)
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= [(x+2)(x^2-2x+4)/(x+2)(x-2)]
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Cancel the (x+2) and (x-2) and (x^2-2x+4) factors to get:
= [1*1/1*1]
= 1/(x-2)
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cheers,
Stan H.
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