Posted by Kelli on Friday, January 4, 2013 at 7:52am.
try factors of 12
on the first try, I got f(1) = 0
so x-1 is a factor
by long algebraic or by synthetic division , I got
x^4+2x^3-7x^2-8x+12
= (x-1)(x^3 + 3x^2 - 4x - 12)
grouping the x^3 + 3x^2 - 4x - 12
= x^2(x+3) - 4(x+3)
= (x+3)(x^2-4)
= (x+3)(x+2)(x-2)
so
x^4+2x^3-7x^2-8x+12
= (x-1)(x-2)(x+2)(x+3)
(had I been patient, I would have found
f(2) = 0, f(-3)=0 and f(-2) = 0 as well)
the zeros are
1, 2, -2, -3
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