# Thread: Factor a cubic without using CAS

1. ## Factor a cubic without using CAS

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

2. ## Re: Factor a cubic without using CAS

Originally Posted by MSUMathStdnt
How do you factor $24x^3-96x^2+102x-30=6(2x-5)(x-1)(2x-1)$ 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:
$4x^3 - 16x^2 + 17x - 5 = 0$

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

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

3. ## Re: Factor a cubic without using CAS

Originally Posted by MSUMathStdnt
How do you factor $24x^3-96x^2+102x-30=6(2x-5)(x-1)(2x-1)$ without a computer? (Obviously, I used a computer).
If we just start with $24x^3-96x^2+102x-30$, look for nice roots.

$x=1$ sort of pops out. So $(x-1)$ 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.