1. ## Factorization method

$\displaystyle 20x^3-68x^2+69x-18=0$ Other than trial and error method... is there any way to factorize this faster?

2. Originally Posted by cloud5
$\displaystyle 20x^3-68x^2+69x-18=0$ Other than trial and error method... is there any way to factorize this faster?
The fastest method is to use a Computer Algebra System...

The one I used got

$\displaystyle 20x^3 - 68x^2 + 69x - 18 =\pm(2x - 3)^2(5x - 2)$.

If you were going to use the Factor and Remainder Theorems, you would need to substitute $\displaystyle x = \frac{2}{5}$ or $\displaystyle x = \frac{3}{2}$ and find that the polynomial equals 0.

3. Originally Posted by Prove It
The fastest method is to use a Computer Algebra System...

The one I used got

$\displaystyle 20x^3 - 68x^2 + 69x - 18 = {\color{red}\pm} (2x - 3)^2(5x - 2)$.

If you were going to use the Factor and Remainder Theorems, you would need to substitute $\displaystyle x = \frac{2}{5}$ or $\displaystyle x = \frac{3}{2}$ and find that the polynomial equals 0.
$\displaystyle \pm$ ?

4. Originally Posted by mr fantastic
$\displaystyle \pm$ ?
I thought it was weird too, but that's what it said...

I think it should say

$\displaystyle [\pm(2x - 3)]^2(5x - 2)$.