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December 20, 2014

December 20, 2014

Posted by **Joe** on Tuesday, October 12, 2010 at 1:44am.

im needing the complex, rational, irrational, and all of the zero's

- math -
**Reiny**, Tuesday, October 12, 2010 at 8:09amI tried synthetic division and fortunaltely it worked nicely for x = 1,2, and 3

so we know that (x-1)(x-2)(x-3) are factors

so we can "reason" out what the fourth factor has to be

at the front so far we have x^3 but we need 2x^4, so the start of the missing factor is 2x

at the end so far we have (-1)(-2)(-3) or -6 so we need +1 at the end of our factor

missing factor is (2x+1)

2x^4-11x^3+16x^2-x-6 = (x-1)(x-2)(x-3)(2x+1)

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