How do you find all the real and complex roots of p(x) = x^5 + 4x^4 + 6x^3 + 4x^2 + 5x?

Jan 11, 2016

The zeros are $x = 0$, $x = \pm i$, $x = - 2 \pm i$

Explanation:

$p \left(x\right) = {x}^{5} + 4 {x}^{4} + 6 {x}^{3} + 4 {x}^{2} + 5 x$

First separate out the common factor $x$ to find:

$p \left(x\right) = x \left({x}^{4} + 4 {x}^{3} + 6 {x}^{2} + 4 x + 5\right)$

Note that $x = 0$ is one of the zeros of $p \left(x\right)$.

Next, let $t = x + 1$

This is called a Tschirnhaus transformation. This kind of simple Tschirnhaus transformation can be used normally to eliminate the ${x}^{3}$ term, but in the case of this particular quartic it helps a lot more.

Then ${t}^{4} = {x}^{4} + 4 {x}^{3} + 6 {x}^{2} + 4 x + 1$

So ${x}^{4} + 4 {x}^{3} + 6 {x}^{2} + 4 x + 5 = {t}^{4} + 4$

This has zeros when ${t}^{4} = - 4$, that is when ${t}^{2} = \pm \sqrt{- 4} = \pm 2 i$

That is when $t = \pm 1 \pm i$

If $t = 1 \pm i$ then $x = t - 1 = \pm i$

If $t = - 1 \pm i$ then $x = t - 1 = - 2 \pm i$