Question

How do you find all the zeros of `f(x) = x^4 - 9x^3 + 24x^2 - 6x - 40`?

Answer 1

`x = -1`, `x = 4`, `x = 3+-i`

Explanation:

First notice that if you reverse the signs of the coefficients on the terms of odd degree then the sum is `0`. That is:

`1+9+24+6-40 = 0`

So `f(-1) = 0` and `(x+1)` is a factor:

`x^4-9x^3+24x^2-6x-40 = (x+1)(x^3-10x^2+34x-40)`

Let `g(x) = x^3-10x^2+34x-40`

By the rational root theorem, any rational zeros of `g(x)` must be expressible in the form `p/q` for integers `p`, `q` with `p` a divisor of the constant term `-40` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-2`, `+-4`, `+-5`, `+-8`, `+-10`, `+-20`, `+-40`

In addition, note the `g(-x) = -x^3-10x^2-34x-40` has all negative coefficients. So there are no negative zeros.

So the only possible rational zeros of `g(x)` are:

`1, 2, 4, 5, 8, 10, 20, 40`

We find:

`g(4) = 64-160+136-40 = 0`

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

`x^3-10x^2+34x-40`

`=(x-4)(x^2-6x+10)`

The remaining quadratic factor has negative discriminant, so its zeros are Complex, but we can factor it as a difference of squares:

`x^2-6x+10`

`=(x-3)^2-9+10`

`=(x-3)^2+1`

`=(x-3)^2-i^2`

`=((x-3)-i)((x-3)+i)`

`=(x-3-i)(x-3+i)`

So the remaining zeros are:

`x = 3+-i`