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April 21, 2014

April 21, 2014

Posted by **Vic** on Friday, February 18, 2011 at 8:38pm.

- Algebra 2 -
**drwls**, Saturday, February 19, 2011 at 1:49pmIf there is a rational root, it would be +or- 1,2,3,4,6,8 or 12.

x = -2 works, so x+2 is a factor.

So, after one factoring step, and a bit of polynomial long division, you have

(x+2)(x^3 -x^2 -6x +18) = 0

Next, use the rational root theorm on the cubic factor. You will find that -3 is a root, making (x+3) a factor.

So now you have

x^4+x^3-8x^2+6x+36

(x+2)(x+3)(x^2 -4x +6)

The remaining two factors are complex conjugates.

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