# How do you find all the zeros of f(x)=x^5+4x^4+5x^3?

Feb 26, 2016

$f \left(x\right) = {x}^{5} + 4 {x}^{4} + 5 {x}^{3} = {x}^{3} \left(x + 2 - i\right) \left(x + 2 + i\right)$

Hence the zeros of $f \left(x\right)$ are $x = 0$ and $x = - 2 \pm i$.

#### Explanation:

First separate out the common factor ${x}^{3}$.

Then complete the square and use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = x + 2$ and $b = i$, as follows:

$f \left(x\right) = {x}^{5} + 4 {x}^{4} + 5 {x}^{3}$

$= {x}^{3} \left({x}^{2} + 4 x + 5\right)$

$= {x}^{3} \left({x}^{2} + 4 x + 4 + 1\right)$

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

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

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

Hence the zeros of $f \left(x\right)$ are $x = 0$ and $x = - 2 \pm i$.