# Thread: Factorization problem

1. ## Factorization problem

I have a problem factorizing that equation 2x^3-5x^2-6x+8 until now I've found the root x=2 and x=-4/3 and I'm stuck there

Thanks for the help

2. ## Re: Factorization problem

Originally Posted by sebastien
I have a problem factorizing that equation 2x^3-5x^2-6x+8 until now I've found the root x=2 and x=-4/3 and I'm stuck there

Thanks for the help
It doesn't factorise... And what you have listed are not roots.

3. ## Re: Factorization problem

made a mistake my equation is 3x^3-5x^2-6x+8 and not the other one

4. ## Re: Factorization problem

Originally Posted by sebastien
made a mistake my equation is 3x^3-5x^2-6x+8 and not the other one
Well, like you have found, x = 2 is a root, which means (x - 2) is a factor.

Long dividing gives

\displaystyle \displaystyle \begin{align*} (x - 2)(3x^2 + x - 4) \end{align*}

and factorising the resulting quadratic gives

\displaystyle \displaystyle \begin{align*} (x - 2)(3x^2 + x - 4) &= (x - 2)(3x^2 - 3x + 4x - 4) \\ &= (x - 2)[3x(x - 1) + 4(x - 1)] \\ &= (x - 1)(x - 2)(3x +4) \end{align*}