can someone tell me how to factor this? (x-3)(x-1)(x+1)(x+3)+64 thanks.
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Originally Posted by 21_knip can someone tell me how to factor this? (x-3)(x-1)(x+1)(x+3)+64 thanks. If you expand and simplify (and there's an easy way to do this by hand) you get $\displaystyle x^4 - 10x^2 + 73$. This can be factorised as two irreducible real quadratics factors and four complex linear factors otherwise.
Last edited by mr fantastic; Jan 13th 2009 at 08:29 PM. Reason: Fixed typo
is this the final answer when it is said to factor completely? ur answer, is it still factorable?
Originally Posted by 21_knip is this the final answer when it is said to factor completely? ur answer, is it still factorable? Is the quartic I have given yu to be factorised over real or complex numbers? Start by thinking how to factorise into two irreducible quadratics.
if i changed 64 to 16 what would be the answer? (x-3)(x-1)(x+1)(x+3)+16 my answer is (x^2-3)(x^2-5) is this correct?
Originally Posted by 21_knip if i changed 64 to 16 what would be the answer? (x-3)(x-1)(x+1)(x+3)+16 my answer is (x^2-3)(x^2-5) is this correct? No. The quartic becomes $\displaystyle x^4 - 10x^2 + 25$ and this has a very simple factorisation (think perfect square) ....
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