Hi,Quote:

Originally Posted byDBA in orginal post

sorry I made a mistake...

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.

Sorry about that

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- Aug 25th 2009, 01:26 PMDBAFactoring with cubesQuote:

Originally Posted by**DBA in orginal post**

sorry I made a mistake...

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.

Sorry about that - Aug 25th 2009, 02:29 PMpickslides
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.

Therefore this is not correct. The term you have here is of order $\displaystyle x^4$

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

You need to apply this theorem to factor a cubic of this nature. - Aug 27th 2009, 12:39 AMPQR$\displaystyle x^3-3x^2-4x+12 = 0$

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

$

$\displaystyle f(-2) = 0$

**∴**$\displaystyle x^3-3x^2-4x+12$**≡**$\displaystyle (x+2)(ax^2+bx+c)$

$\displaystyle x^3-3x^2-4x+12$**≡**$\displaystyle ax^3+bx^2+cx+2ax^2+2bx+2c$

$\displaystyle x^3-3x^2-4x+12$**≡**$\displaystyle ax^3+bx^2+2ax^2+cx+2bx+2c$

$\displaystyle x^3-3x^2-4x+12$**≡**$\displaystyle ax^3+(b+2a)x^2+(c+2b)x+2c$

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

**∴**$\displaystyle x^3-3x^2-4x+12$**≡**$\displaystyle (x+2)(x^2-5x+6)$

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

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

$\displaystyle f(2) = 0$

∴ $\displaystyle x^2-5x-4$ ≡ $\displaystyle (x-2)(ax+b)$

$\displaystyle x^2-5x-4 $ ≡ $\displaystyle ax^2+bx-2ax-2b$

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

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

∴ $\displaystyle x^2-5x-4$ ≡ $\displaystyle (x-2)(x-3)$

∴ $\displaystyle x^3-3x^2-4x+12$ ≡ $\displaystyle (x+2)(x-2)(x-3)$