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I saw this problem on sosmath.com - http://www.sosmath.com/algebra/pfrac/pfrac.html
x^3 - 4x^2 + 5
and it's suppose to factor out to (x+1)(x^2-5x+5) and I've been sitting here trying to get it to that form but I only end up with (x^2+5)(x-4). Can someone please shed some light to how this can be factored out to (x+1)(x^2-5x+5). Thanks!
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x = -1 can be spotted as a root of the equation quite easily as the coeffecients of the x^2 term and the 5 sum to 1.
(-1)^3 - 4(-1)^2 + 5 = -1 - 4 + 5 = 0
so you have:
(x+1)(x^2 + bx + c) (coef. of x^2 is 1 since it's multiplied with x to give x^3)
x^3 + (b+1)x^2 + (c+b)x + c = x^3 - 4x^2 + 5
which gives immediately, b+1 = -4 -> b = -5,, c+b = 0 -> c = 5 (or just c = 5 immediately from other coef.)
(x+1)(x^2 - 5x + 5)
Last edited by luca-deltodesco (2009-10-25 20:32:46)
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Thanks Luca for your response, but I'm still confused to how (x+1)(x^2-5x+5) was factored out from x^3-4x^2+5.
Is (x+1)(x^2+bx+x) a identity that has to be recognized through x^3-4x^2+5?
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Hi ccmint;
The fact that (x+1)(x^2-5x+5) are factors of x^3-4x^2+5 means that when multiplied:
[/math](x+1)(x^2-5x+5) = x^3-4x^2+5[/math]
When you know a root r, you can factor out an (x-r). Since Luca found that x = -1 is a root, one factor is (x+1). Now just divide x^3-4x^2+5 by x+1 and you will get (x^2-5x+5).
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thanks Bobby and Luca. It all makes sense now.
Have a great day, John
Last edited by ccmint (2009-10-26 05:04:52)
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Hi ccmint;
Thanks, we are glad to help. Have a better day!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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