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

Apr 16, 2016

${x}^{4} - 4 {x}^{3} - 20 {x}^{2} + 48 x = x \left(x - 2\right) \left(x - 6\right) \left(x + 4\right)$

#### 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:

$f \left(x\right) = {x}^{3} - 4 {x}^{2} - 20 x + 48$

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

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 8 , \pm 12 , \pm 16 , \pm 24 , \pm 48$

We find $f \left(2\right) = 8 - 16 - 40 + 48 = 0$

So $x = 2$ is a zero and $\left(x - 2\right)$ a factor:

${x}^{3} - 4 {x}^{2} - 20 x + 48 = \left(x - 2\right) \left({x}^{2} - 2 x - 24\right)$

To factor the remaining quadratic, note that $6 - 4 = 2$ and $6 \times 4 = 24$, hence:

${x}^{2} - 2 x - 24 = \left(x - 6\right) \left(x + 4\right)$

Putting it all together:

${x}^{4} - 4 {x}^{3} - 20 {x}^{2} + 48 x = x \left(x - 2\right) \left(x - 6\right) \left(x + 4\right)$