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I did the P/Q thing and used synthetic division, that brought me to
(x-2) (x[cubed] + 6x[squared] + 9x + 4

I've absolutely no idea where to go from here. Help? I know I need to factor down the equation so I can fit it into the quadratic formula, then find the roots from there but I've no idea how to get past the x cubed.

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Re: Rational Root Theorem? -
February 10th 2009, 02:43 AM

Well I guessed substitutes and found than -1 makes
x³ - 6x² +9x + 4 equal 0 so this makes (x+1) a factor of x³ - 6x² +9x + 4 so dividing through by (x+1) gives x² + 7x + 4 = 0 which in turn has roots of [-7+sqrt(33)]/2 or [-7-sqrt(33)]/2