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

Feb 27, 2016

Factors of f(x)=x^4−4x^3−20x^2+48x) are $x \left(x - 2\right) \left(x + 4\right) \left(x - 6\right)$ and zeros of $f \left(x\right)$ are $\left\{0 , 2 , - 4 , 6\right\}$

#### Explanation:

As there is no independent in f(x)=x^4−4x^3−20x^2+48x and hence$x$ is factor of $f \left(x\right)$ and hence ) is first zero among the zeros of $f \left(x\right)$. Factorizing $f \left(x\right)$

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

Now identify factors of $48$ i.e. $\left\{1 , - 1 , 2 , - 2 , 3 , - 3 , 4 , - 4 , 6 , - 6 , \ldots\right\}$, who make $\left({x}^{3} - 4 {x}^{2} - 20 x + 48\right) = 0$. As $x = 2$, makes $f \left(x\right) = 0$, and hence $\left(x - 2\right)$ is a factor of $f \left(x\right)$ and $2$ is another zero of $f \left(x\right)$.

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

Now $\left({x}^{2} - 2 x - 24\right)$, can be easily factorized as $- 24$ can be exactly factorized into $- 6$ and $4$, and hence factorization can proceed as under.

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

Hence complete factors of f(x)=x^4−4x^3−20x^2+48x) are $x \left(x - 2\right) \left(x + 4\right) \left(x - 6\right)$ and zeros of $f \left(x\right)$ are $\left\{0 , 2 , - 4 , 6\right\}$