Question

How do you find all the zeros of `x^4 - 4x^3 - 20x^2 + 48x`?

Answer 1

`x^4-4x^3-20x^2+48x = x(x-2)(x-6)(x+4)`

Explanation:

Firstly, since all of the factors are divisible by `x`, there is a zero `x=0`.

Dividing through by `x` we get a cubic:

`f(x) = x^3-4x^2-20x+48`

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `48` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-3, +-4, +-6, +-8, +-12, +-16, +-24, +-48`

We find `f(2) = 8-16-40+48 = 0`

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

`x^3-4x^2-20x+48 = (x-2)(x^2-2x-24)`

To factor the remaining quadratic, note that `6-4 = 2` and `6 xx 4 = 24`, hence:

`x^2-2x-24 = (x-6)(x+4)`

Putting it all together:

`x^4-4x^3-20x^2+48x = x(x-2)(x-6)(x+4)`