# Factor a cubic without using CAS

• Dec 10th 2012, 10:05 AM
MSUMathStdnt
Factor a cubic without using CAS
How do you factor $\displaystyle 24x^3-96x^2+102x-30=6(2x-5)(x-1)(2x-1)$ without a computer? (Obviously, I used a computer).
• Dec 10th 2012, 10:24 AM
topsquark
Re: Factor a cubic without using CAS
Quote:

Originally Posted by MSUMathStdnt
How do you factor $\displaystyle 24x^3-96x^2+102x-30=6(2x-5)(x-1)(2x-1)$ without a computer? (Obviously, I used a computer).

Are you aware of the Rational Root Theorem?

For convenience I'm going to divide the polynomial by 6:
$\displaystyle 4x^3 - 16x^2 + 17x - 5 = 0$

Possible rational roots to this polynomial equation are:
$\displaystyle \pm \left ( 5,~\frac{5}{2},~\frac{5}{4},~1,~\frac{1}{2},~\frac {1}{4} \right )$

That gives you 12 possible rational roots that you can check.

If you don't know that one, then you're probably stuck with Cardano's method. (Cardano's method is about 1/2 way down the page.)

-Dan
• Dec 10th 2012, 10:26 AM
Plato
Re: Factor a cubic without using CAS
Quote:

Originally Posted by MSUMathStdnt
How do you factor $\displaystyle 24x^3-96x^2+102x-30=6(2x-5)(x-1)(2x-1)$ without a computer? (Obviously, I used a computer).

If we just start with $\displaystyle 24x^3-96x^2+102x-30$, look for nice roots.

$\displaystyle x=1$ sort of pops out. So $\displaystyle (x-1)$ must be a factor.

Once we have one factor then by division we can reduce that to a quadratic which can factored.

I understand that it is not always possible to find a root that easily.