SOLUTION: solve: x^3-x^2+4*x-4=0 thanks, chris

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Question 50682: solve:
x^3-x^2+4*x-4=0
thanks,
chris
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
This is factoring by grouping. Group the first two and last two terms together.

x%5E3-x%5E2%2B4%2Ax-4=0
x%5E2%28x-1%29+%2B+4%28x-1%29+=+0

Notice that there is a common factor of (x-1):
+%28x-1%29+%28x%5E2%2B4%29+=+0
x=1 or x%5E2+=+-4

Are you solving with complex numbers or real numbers only? If real numbers only, then the only answer you have to worry about is x=1. End of problem, since real numbers squared cannot equal a negative.

However, if complex numbers are allowed, then
x%5E2+=+-4 gives two solutions:
x=+sqrt%28-4%29, x=-sqrt%28-4%29+
x=+2i, x=+-2i in addition to x=+1.

R^2 at SCC