# I forgot how to factor cubic equations?

• yurkler

#### yurkler

the original equation is 4x^2-6x^2-12x+9=0

I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0

Apply the rational root test to find a possible root.

The possible rational roots will have the form p/q where p divides 9 and where q divides 4. So what are the possible rational roots?? Is one of these possibilities an actual root?

the original equation is 4x^2-6x^2-12x+9=0

I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0
Ah, this looks like the "factoring by grouping" method, which, unfortunately, doesn't work for ALL cubic polynomials. The rational root test, as micromass mentioned, is your best bet.

the original equation is 4x^2-6x^2-12x+9=0

I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0

Do you mean 4x^3 ?

The equation you posted is not a cubic; it is a quadratic.

4 x^2 - 6 x^2 . . .

equals -2 x^2 . . .

Why didn't you combine the x^2 terms?

In any case, equations exist for solving a quadratic or cubic equation. Just do a search; I am sure you will find the appropriate formula.

You mean 4x3-6x2-12x+9=0?

The roots of a polynomial must be a factor of the product of the first term and the last term. Therefore, we have to try all the factors for 4*9=36. There's this really convenient method for factoring polynomials like this. It's kind of like long division, but not really. Let's try x=4 for example. Then we set up the polynomial division like this:

2 | 4, -6, -12, 9
..|.....

First step. Drop the first term (from the left) to the right of the bar.

2 | 4, -6, -12, 9
..|.....
...4

Next, multiply that number by 2, and put it below the next entry.

2 | 4, -6, -12, 9
..|...8...
...4

2 | 4, -6, -12, 9
..|...8...
...4...2

Repeat

2 | 4, -6, -12, 9
..|...8...4...
...4...2...-8

2 | 4, -6, -12, 9
..|...8...4..-16
...4...2...-8..-7

The process I described reveals

4x3-6x2-12x+9 = 4x2-2x-8-7/(x-2)

The last term there, -7/(x-2) is called the remainder term. Since there is a remainder term, 2 is not a root. You just have to keep testing roots until you get one without a remainder term.

You mean 4x3-6x2-12x+9=0?

The roots of a polynomial must be a factor of the product of the first term and the last term. Therefore, we have to try all the factors for 4*9=36. There's this really convenient method for factoring polynomials like this. It's kind of like long division, but not really.

(Snip)

What about just trying to substitute a possible root in the original expression, instead of doing a polynomial division?

You could do that and then use the long division to find the coefficients of the remaining quadratic.