Question

How to factor cubic trinomials? x^3-7x-6

Answer 1

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

Explanation:

You could solve this by plotting the equation and inspecting where the roots are:
graph{x^3-7x-6 [-5, 5, -15, 5]}

We can see there appear to be roots in the areas of `x=-2,-1,3`, if we try these we see this is indeed a factorisation of the equation:

`(x-3)(x+1)(x+2)=(x-3)(x^2+3x+2)=x^3-7x-6`

Answer 2

Use the rational roots theorem to find possible roots, try each to find roots `x=-1` and `x=-2` hence factors `(x+1)` and `(x+2)` then divide by these to find `(x-3)`

`x^3-7x-6 = (x+1)(x+2)(x-3)`

Explanation:

Find roots of `x^3-7x-6 = 0` and hence factors of `x^3-7x-6`.

Any rational root of a polynomial equation in standard form is of the form `p/q`, where `p`, `q` are integers, `q != 0`, `p` a factor of the constant term and `q` a factor of the coefficient of the term of highest degree.

In our case `p` must be a factor of `6` and `q` a factor of `1`.

So the only possible rational roots are: `+-1`, `+-2`, `+-3` and `+-6`.

Let `f(x) = x^3-7x-6`

`f(1) = 1-7-6 = -12`
`f(-1) = -1+7-6 = 0`
`f(2) = 8-14-6 = -12`
`f(-2) = -8+14-6 = 0`

So `x = -1` is a root of `f(x) = 0` and `(x+1)` a factor of `f(x)`.
`x=-2` is a root of `f(x) = 0` and `(x+2)` a factor of `f(x)`.

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

Divide `f(x)` by the factors we've found so far to find:

`x^3-7x-6 = (x^2+3x+2)(x-3)`

Actually you can deduce the `x` and the `-3` simply by looking at what you need to multiply `x^2` and `2` by to get `x^3` and `-6`.

So the complete factorisation is:

`x^3-7x-6 = (x+1)(x+2)(x-3)`