SOLUTION: We have been working on factoring trinomials and the problem I don't know how to solve is (x^2+4x-4)^2=64. I think what we're supposed to do is divide the exponent off both sides b

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Question 354730: We have been working on factoring trinomials and the problem I don't know how to solve is (x^2+4x-4)^2=64. I think what we're supposed to do is divide the exponent off both sides but I'm not sure. Were I get stuck is you're left with x^2+4x-4 on one side and that doesn't facter into two binomials. HELP ME!
Found 2 solutions by scott8148, Edwin McCravy:
Answer by scott8148(6628)   (Show Source): You can put this solution on YOUR website!
taking square root ___ x^2 + 4x - 4 = ±8

x^2 + 4x - 4 = 8 ___ x^2 + 4x - 12 = 0

x^2 + 4x - 4 = -8 ___ x^2 + 4x + 4 = 0

now factor both ...



Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!

We use the principle of square roots:

Take away the square on the right, take ± square roots of the right side





Make two equations, one with the + and one with the -

 and 

Solving the first one



Get 0 on the right:



Factor the left side by thinking of two whole numbers which
have product 12 and difference 4.  They are 6 and 2.  So
we write

  

Then we put in the signs so that if we were to FOIL it out
we would get the middle term  in the middle.



Then we use the zero-factor principle:

set each factor = 0

 gives solution 

 gives solution 

-------------------

Solving the second one



Get 0 on the right:



Factor the left side by thinking of two whole numbers which
have product 4 and sum 4.  They are 2 and 2.  So
we write

  

Then we put in the signs so that if we were to FOIL it out
we would get the middle term  in the middle.



Then we use the zero-factor principle:

set each factor = 0

 gives solution 

The second factor is the same, so we do not get an additional solution.

Checking in the original equation:

Checking 








Checking 








Checking 








They are all solutions.

Edwin

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