Question

How do you factor `z^3-z^2-13z+4`?

Answer 1

`z^3-z^2-13z+4 = (z-4)(z + 3/2 - sqrt(5)/2)(z + 3/2 + sqrt(5)/2)`

Explanation:

By the rational root theorem, any rational zeros must be expressible in the form `p/q` for some integers `p` and `q` with `p` a divisor of the constant term `4` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-2`, `+-4`

Let `f(z) = z^3-z^2-13z+4`

We find `f(4) = 64-16-52+4 = 0`, so `z=4` is a zero and `(z-4)` is a factor:

`z^3-z^2-13z+4 = (z-4)(z^2+3z-1)`

We can factor the remaining quadratic factor by completing the square:

`z^2+3z-1`

`= (z+3/2)^2-9/4+1`

`= (z+3/2)^2-5/4`

`= (z+3/2)^2-(sqrt(5)/2)^2`

`= (z + 3/2 - sqrt(5)/2)(z + 3/2 + sqrt(5)/2)`

Putting it all together:

`z^3-z^2-13z+4 = (z-4)(z + 3/2 - sqrt(5)/2)(z + 3/2 + sqrt(5)/2)`