Question

How do you factor `y= 6x^3+13x-14x+3` ?

Answer 1

Use the rational root theorem to help find the first root and factor, then an AC method to factor the remaining quadratic to find:

`6x^3+13x^2-14x+3 = (3x-1)(2x-1)(x+3)`

Explanation:

I will guess that `13x` should have been `13x^2` in the question.

Let `f(x) = 6x^3+13x^2-14x+3`

By the rational root theorem, any rational zeros of `f(x)` are expressible in lowest terms in the form `p/q`, where `p` and `q` are integers, `p` a divisor of the constant term `3` and `q` a divisor of the coefficient `6` of the leading term.

That means that the only possible rational roots are:

`+-1/6`, `+-1/3`, `+-1/2`, `+-1`, `+-3/2`, `+-3`

Let's try some of these:

`f(1/6) = 6/216+13/36-14/6+3`

`=(1+13-84+108)/36 = 38/36 = 19/18`

`f(-1/6) = (-1+13+84+108)/36 = 204/36 = 17/3`

`f(1/3) = 6/27+13/9-14/3+3`

`=(2+13-42+27)/9 = 0`

So `x=1/3` is a zero and `(3x-1)` is a factor:

`6x^3+13x^2-14x+3 = (3x-1)(2x^2+5x-3)`

Then use an AC method to help factor `2x^2+5x-3`.

Look for a pair of factors of `AC = 2*3 = 6` that differ by `B=5`.

The pair `6, 1` works, hence:

`2x^2+5x-3`

`= 2x^2+6x-x-3`

`= (2x^2+6x)-(x+3)`

`= 2x(x+3)-1(x+3)`

`= (2x-1)(x+3)`

Putting it all together:

`6x^3+13x^2-14x+3 = (3x-1)(2x-1)(x+3)`