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July 25, 2014

July 25, 2014

Posted by **kaley** on Thursday, March 1, 2012 at 10:27pm.

F(x)= x^4+4x^3-6x^2-36x-27

- Algebra 2 -
**drwls**, Thursday, March 1, 2012 at 11:02pmThere are websites that solve such root-finding questions automatically, but they are not very instructive.

If there are integer roots, they must be even divisors of 27: +/- 1, 3, 9 or 27.

That is a consequence of the "rational roots theorem", which you should learn.

One such root is -1, so x+1 is a factor.

The other factor is

(x^4+4x^3-6x^2-36x-27)/(x+1)

= x^3 +3x^2 -9x -27

(obtained with polynomial long division)

x = 3 is clearly another root, so (x-3) is another factor. Divide the cubic by (x-3) and you get

x^2 + 6x +9 = 0

which can be factored to give

(x+3)^2 = 0

That means x = -3 is a double root.

The roots are -3, +3 and -1

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