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December 20, 2014

December 20, 2014

Posted by **courtney** on Wednesday, March 27, 2013 at 12:06am.

f(x)=4x^4+9x^3+30x^2+63x+14

I cant even find sample problems to help me figure this out. help me please?

- math -
**drwls**, Wednesday, March 27, 2013 at 7:37amTry roots of the form x = +/- p/q, where p is an integer factor of 14 (1,2,7,14) and q is a factor of 4 (1,2,4)

You will have to chooose a negative value of x to get a negative or zero value of f(x). Try x = -2/1 = -2

f(x) = 64 - 72 + 120 -126 + 14 = 0

So x = 2 is a solution . You can get another real root by dividing

(4x^4+9x^3+30x^2+63x+14)/(x+2)

and solving the remaining cubic.

The third and fourth roots are complex conjugates.

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