Question

How do you find all zeros with multiplicities of `f(x)=x^4+2x^3-12x^2-40x-32`?

Answer 1

`f(x)` has zeros `4` (with multiplicity `1`) and `-2` with multiplicity `3`.

Explanation:

Given:

`f(x) = x^4+2x^3-12x^2-40x-32`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divosor of the constant term `-32` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1, +-2, +-4, +-8, +-16, +-32`

Note also that the signs of the coefficients are in the pattern `+ + - - -`. By Descartes' Rule of Signs, sinc this has one change, we can deduce that `f(x)` has one positive real zero. So let us look for that one first.

`f(4) = 4^4+2(4^3)-12(4^2)-40(4)-32 = 256+128-192-160-32 = 0`

So `x=4` is a zero and `(x-4)` a factor:

`x^4+2x^3-12x^2-40x-32 = (x-4)(x^3+6x^2+12x+8)`

Let us try substituting `-2` in the remaining cubic factor:

`(-2)^3+6(-2)^2+12(-2)+8 = -8+24-24+8 = 0`

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

`x^3+6x^2+12x+8 = (x+2)(x^2+4x+4)`

The remaining quadratic factor is a perfect square trinomial, as we might spot by noticing `144 = 12^2` and similarly:

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

So `x=-2` is a zero twice more.

graph{x^4+2x^3-12x^2-40x-32 [-10, 10, -160, 80]}