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July 29, 2014

July 29, 2014

Posted by **Armando** on Monday, May 30, 2011 at 12:28am.

f(x) = x^3 + 2x^2 - 6x + 8

- find real and complex 0 -
**Reiny**, Monday, May 30, 2011 at 8:51amf(±1) ≠ 0

f(±2) ≠ 0

but f(-4) = 0

so x+4 is a factor

I then used synthetic division to get

(x+4)(x^2 - 2x + 2) = 0

the real root is x = -4

Use the quadratic formula to get the two complex roots of

1+i and 1-i

- find real and complex 0 -
**Mgraph**, Monday, May 30, 2011 at 9:03amf(x)=x^3+4x^2-2x^2-8x+2x+8=(x^3+4x^2)-

(2x^2+8x)+(2x+8)=x^2(x+4)-2x(x+4)+2(x+4)=

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

x+4=0 => x=-4

or

x^2-2x+2=0 => x=1+i or x=1-i

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