# Math Help - Factoring with cubes

1. ## Factoring with cubes

Originally Posted by DBA in orginal post
Hi,
- I stuck with factoring of the following:

x^3-2x^2-4x+12

I thought I could use grouping like

(x^3-2x^2)- (4x-12)=

x^2(x-2) - 4(x-3)
+
it should be

- but this does not give me a common factor, so I still have 3 terms
(x^2-4) (x-2) (x-3)
Hi,

it should be

x^3-3x^2-4x+12

where I than get different terms
(x^3-3x^2)- (4x-12)=

x^2(x-3) - 4(x-3)

and then it works out fine.

2. Originally Posted by DBA
Hi,
I stuck with factoring of the following:

x^3-2x^2-4x+12

I thought I could use grouping like

(x^3-2x^2)- (4x-12)=

x^2(x-2) - 4(x-3)
You can do this but it won't help. When grouping you need the terms inside the brackets to be the same. This is not the case here in what you have done.

Originally Posted by DBA

but this does not give me a common factor, so I still have 3 terms
(x^2-4) (x-2) (x-3)
Therefore this is not correct. The term you have here is of order $x^4$

Do you know the factor theorem? It is, for $P(x)$, if $P(a) = 0$ then $x-a$ is a factor.

You need to apply this theorem to factor a cubic of this nature.

3. $x^3-3x^2-4x+12 = 0$

$[(x-3)x-4]x+12 = 0
$

$f(-2) = 0$

$x^3-3x^2-4x+12$ $(x+2)(ax^2+bx+c)$

$x^3-3x^2-4x+12$ $ax^3+bx^2+cx+2ax^2+2bx+2c$

$x^3-3x^2-4x+12$ $ax^3+bx^2+2ax^2+cx+2bx+2c$

$x^3-3x^2-4x+12$ $ax^3+(b+2a)x^2+(c+2b)x+2c$

$a =1$; $b+2a = -3$, $b = -5$; $c+2b -4$, $c = 6$.

$x^3-3x^2-4x+12$ $(x+2)(x^2-5x+6)$

$x^2-5x-6 = 0$

$(x-5)x-6 = 0$

$f(2) = 0$

$x^2-5x-4$ $(x-2)(ax+b)$

$x^2-5x-4$ $ax^2+bx-2ax-2b$

$x^2-5x-4$ $(a)x^2+(b-2a)x-2b$

$a = 1$; $b-2a$ = $-5$, $b = -3$.

$x^2-5x-4$ $(x-2)(x-3)$

$x^3-3x^2-4x+12$ $(x+2)(x-2)(x-3)$