Posted by **Natasha** on Monday, September 17, 2012 at 11:03pm.

find the real zeros of f(x)=x^4+6x^3-11x^2-24x+28

- pre calculus -
**Reiny**, Monday, September 17, 2012 at 11:33pm
try x=1

f(1) = 1+6-11-24+28 = 0

so x-1 is a factor

do synthetic division

(x^4+6x^3-11x^2-24x+28) = (x-1)(x^3 + 7x - 4 - 28)

x^3 + 7x^2 - 4x - 28

= x^2(x+7) - 4(x+7)

= (x+7)(x^ - 4)

so x^4+6x^3-11x^2-24x+28

= (x-1)(x-2)(x+2)(x+7)

x = 1, -2, 2, -7

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