SOLUTION: 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)

Question 288610: 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)?
I tried out some problems, and figured out the remaining factors should be either (x+1),(x-2) or the remaining factors should be: (x-1),(x+2)........or am I wrong completely wrong?
And, could you please show me the way on HOW to figure this out? Thank you
Found 2 solutions by richwmiller, stanbon:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
The equation that you gave us x^3+5x^2=2x-8 does not have (x+4) as a factor.
I suspect you copied something wrong.
My best guess is that it isn't an equation at all and the = should be +
So here it is
(x^3+5x^2+2x-8)/(x+4)=
x^2+x-2=
(x-1)*(x+2)

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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...
------------------
Please check your problem to see if you posted it correctly.
==========================
Cheers,
Stan H.

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