Friday

April 18, 2014

April 18, 2014

Posted by **phyllys** on Saturday, January 5, 2013 at 10:33am.

- Pre-Cal. -
**Steve**, Saturday, January 5, 2013 at 3:48pmyou know that are all factors of 12. A little regrouping shows you have

f(x) = x^3-3x^2+4x-12

= x^2(x-3) + 4(x-3)

= (x^2+4)(x-3)

so, f(x) = 0 at x = 3, ±2i

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