# Factor completely

• Oct 17th 2012, 07:17 PM
ford2008
Factor completely
Hey guys,

the task: Factor completely (that is, remove common factors first).

a)
x^3 +x^2 -12x

Could somebody please explain how I can factor completely and remove common factors first?
• Oct 17th 2012, 07:21 PM
MarkFL
Re: Factor completely
Observe that all terms have x as a factor, so pull that out first, then you are left with a quadratic...to factor this you need two factors of -12 whose sum is 1. Can you finish?
• Oct 17th 2012, 08:44 PM
ford2008
Re: Factor completely
-3 times 4 = -12 and
the sum of -3+4 = 1
• Oct 17th 2012, 08:46 PM
MarkFL
Re: Factor completely
Yes, so what is the factored form of the given expression?
• Oct 18th 2012, 08:40 AM
ford2008
Re: Factor completely
Factored form should be (x-3)(x+4) = x^2 +4x -3x -12 = x^2 +x -12 = x(x^2 +x -12) = x^3 +x^2 -12x

So the factored form should be just: x(x^2 +x -12) or not???
• Oct 18th 2012, 09:01 AM
Plato
Re: Factor completely
Quote:

Originally Posted by ford2008
Factored form should be (x-3)(x+4) = x^2 +4x -3x -12 = x^2 +x -12 = x(x^2 +x -12) = x^3 +x^2 -12x
So the factored form should be just: x(x^2 +x -12) or not???

Completely factored is $x(x+4)(x-3)~.$
• Oct 18th 2012, 09:34 AM
ford2008
Re: Factor completely
2 other examples:

1.
2x^2 +4x -30

Common factor is 2

so 2(x-3)(x+5) is completely factored???

2.
3x^3 +21x^2 +36x

Common factor is 3x

so 3x(x+3)(x+4) is completely factored???