Hey I need some help with this problem Factor the polynomial x^8-98x^4+1 into two non-constant polynomials with integral coefficients. I want to see steps on how to do this problem thanks.
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Start by letting , then the equation becomes . This doesn't factorise over the integers, but does factorise over the real numbers by completing the square.
I completed the square and got (x-49)^2 +2400 what do I do next?
That can't possibly be right - expand it out and check...
sorry it should be -2400 (x-49)^2 - 2400 = (x-49)(x-49) - 2400 = x^2 -49x - 49x + 2401 -2400 = x^2 - 98x + 1 ok so when completing the square i get (x-49)^2 - 2400 what do I do next?
Write it as and then use the Difference of Two Squares rule to factorise it.
so it is ((x-49)^2 + (20sqrt(6))^2)((x-49)^2 - (20sqrt(6))^2)) what do I do from here?
Last edited by gfbrd; Nov 10th 2012 at 11:59 PM.
Hello, gfbrd! This takes a little Imagination . . . Factor into two non-constant polynomials with integral coefficients.
lol sorry my imagination isn't really as good as yours thanks a lot soroban
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