Question

How do you find all zeros with multiplicities of `f(x)=x^4-9x^2-4x+12`?

Answer 1

`f(x)` has zeros:

`1` with multiplicity `1`

`3` with multiplicity `1`

`-2` with multiplicity `2`

Explanation:

Given:

`f(x) = x^4-9x^2-4x+12`

First note that the sum of the coefficients is `0`.

That is:

`1-9-4+12 = 0`

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

`x^4-9x^2-4x+12 = (x-1)(x^3+x^2-8x-12)`

By the rational roots theorem, any rational zeros of `x^3+x^2-8x-12` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-12` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational zeros of this cubic are:

`+-1, +-2, +-3, +-4, +-6, +-12`

We find:

`color(blue)(3)^3+color(blue)(3)^2-8*color(blue)(3)-12 = 27+9-24-12 = 0`

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

`x^3+x^2-8x-12 = (x-3)(x^2+4x+4)`

The remaining quadratic factor can be recognised as a perfect square trinomial. Like `144=12^2` we find:

`x^2+4x+4 = (x+2)^2`

Hence the other root is `-2` with multiplicity `2`.