I forgot how to factor cubic equations?

[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

 

[micromass]

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?

 

[eumyang]

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.

 

[epenguin]

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 ?

 

[DuncanM]

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.

 

[Harrisonized]

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

Add that column.

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.

 

[I like Serena]

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?

 

[Harrisonized]

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