SOLUTION: factor completely with respect to the intergers x^4-13x^2+36

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Question 59011: factor completely with respect to the intergers x^4-13x^2+36
Found 2 solutions by stanbon, funmath:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
x^4-13x^2+36
=(x^2-9)(x^2-4)
=(x-3)(x+3)(x+2)(x-2)
Cheers,
Stan H.

Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
factor completely with respect to the intergers
x%5E4-13x%5E2%2B36
x%5E4-4x%5E2-9x%5E2%2B36 Replace middle term with two terms that multiply to give you the product of the outer terms: x^4*36=36x^4, but add to give you the middle term:-13x^2. Then factor by grouping.
%28x%5E4-4x%5E2%29%2B%28-9x%5E2%2B36%29
x%5E2%28x%5E2-4%29-9%28x%5E2-4%29
%28x%5E2-4%29%28x%5E2-9%29 Both parenthesese are the differences of perfect squares.
%28x%2B2%29%28x-2%29%28x%2B3%29%28x-3%29
Happy Calculating!!!