Question

How do you factor `13a^3 - 16a - 72`?

Answer 1

`13a^3-16a-72`

`= (a-2)(13a^2+26a+36)`

`= (a-2)(sqrt(13)a+sqrt(13)-sqrt(23)i)(sqrt(13)a+sqrt(13)+sqrt(23)i)`

Explanation:

By the rational root theorem, any rational zeros of this quadratic will be expressible in the form `a=p/q` for integers `p, q` with `p` a divisor of the constant term `-72` and `q` a divisor of the coefficient `13` of the leading term.

So the possible rational zeros are:

`+-1/13, +-2/13, +-3/13, +-4/13, +-6/13, +-8/13, +-9/13, +-12/13, +-1, +-18/13, +-24/13, +-2, +-36/13, +-3, +-6, +-8, +-9, +-12, +-18, +-24, +-36, +-72`

Let `f(a) = 13a^3-16a-72`

We can rule out any of the non-integral fractions, since for such fractions, `13a^3` will have a denominator `13^2` in lowest terms, which will not be possible to cancel out by the other terms.

So try the integer possibilities in sequence.

We find `f(2) = 13*8-16*2-72 = 104-32-72 = 0`

So `a=2` is a zero and `(a-2)` a factor:

`13a^3-16a-72 = (a-2)(13a^2+26a+36)`

The remaining quadratic factor has negative discriminant:

`Delta = 26^2-(4*13*36) = 676-1872 = -1196`

so no linear factors with Real coefficients.

We can factor using Complex coefficients as follows:

`(a-2)(13a^2+26a+36)`

`= (a-2)(13(a^2+2a+1)+23)`

`= (a-2)(13(a+1)^2+23)`

`= (a-2)((sqrt(13)(a+1))^2-(sqrt(23)i)^2)`

`= (a-2)(sqrt(13)(a+1)-sqrt(23)i)(sqrt(13)(a+1)+sqrt(23)i)`

`= (a-2)(sqrt(13)a+sqrt(13)-sqrt(23)i)(sqrt(13)a+sqrt(13)+sqrt(23)i)`