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

May 8, 2016

$x = - 1$, $x = 4$, $x = 3 \pm i$

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 $f \left(- 1\right) = 0$ and $\left(x + 1\right)$ is a factor:

${x}^{4} - 9 {x}^{3} + 24 {x}^{2} - 6 x - 40 = \left(x + 1\right) \left({x}^{3} - 10 {x}^{2} + 34 x - 40\right)$

Let $g \left(x\right) = {x}^{3} - 10 {x}^{2} + 34 x - 40$

By the rational root theorem, any rational zeros of $g \left(x\right)$ must be expressible in the form $\frac{p}{q}$ 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:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 5$, $\pm 8$, $\pm 10$, $\pm 20$, $\pm 40$

In addition, note the $g \left(- x\right) = - {x}^{3} - 10 {x}^{2} - 34 x - 40$ has all negative coefficients. So there are no negative zeros.

So the only possible rational zeros of $g \left(x\right)$ are:

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

We find:

$g \left(4\right) = 64 - 160 + 136 - 40 = 0$

So $x = 4$ is a zero and $\left(x - 4\right)$ a factor:

${x}^{3} - 10 {x}^{2} + 34 x - 40$

$= \left(x - 4\right) \left({x}^{2} - 6 x + 10\right)$

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

${x}^{2} - 6 x + 10$

$= {\left(x - 3\right)}^{2} - 9 + 10$

$= {\left(x - 3\right)}^{2} + 1$

$= {\left(x - 3\right)}^{2} - {i}^{2}$

$= \left(\left(x - 3\right) - i\right) \left(\left(x - 3\right) + i\right)$

$= \left(x - 3 - i\right) \left(x - 3 + i\right)$

So the remaining zeros are:

$x = 3 \pm i$