<PRE><B>Question 288610</B>
I have a question about the remaining factors.
If you have an equation of x^3+5x^2=2x-8 how do you find the remaining factors, if you know that one factor of the equation is (x+4)?
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Rearrange the equation:
x^3+5x^2-2x+8 = 0
Divide the cubic by x+4 using long division or synthetic division:
-4)....1....5....-2...8
.......1....1....-6...|32

This indecates that (x+4) is NOT a factor of the cubic.
I graphed the cubic and found its only Real Number zero
is around -5.610606...

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Please check your problem to see if you posted it correctly.
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Cheers,
Stan H.
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