x-intercepts of x^4-4x^3+x^2+6x

me3po

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Nov 24, 2006
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Help! I have the equation x^4-4x^3+x^2+6x and I need to find x-intercepts.

I start by factoring out an x so that I have x(x^3-4x^2+x+6)

and then I hit a brick wall, I just can not get the rest to factor.

If someone could give me a nudge in the right direction I would be very greatful, for it is almost 5am and I have been up all night trying to finish this homework.

~R~
 
x(x^3-4x^2+x+6) = 0

You must hunt. The part in the parentheses is an odd degree, so there IS a solution.

DeCartes' Rule of Signs suggests 2 or 0 greater than zero and 1 kess than zero. This is good news. There IS one less than zero.

The constant term suggests only a few possibilities if the negative solution is Rational. -1, -2, -3, -6 Try them. As it turns out, (-1)^3-4(-1)^2+(-1)+6 = 0.

If we can't find a rational root, there are other ways to hunt.
 
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