Question

explain how to find each root, with its multiplicity, for the polynomial y=x⁵-27x²?

Answer 1

$y=x^5-27x^2$

In order to find each root we need to know when:

$y=0$

so:

$\begin{gathered} x^5-27x^2=0 \\ \end{gathered}$

Factor x² out of x⁵- 25x²:

$\begin{gathered} x^2(x^3-27) \\ where \\ x^3-27=x^3-3^3 \end{gathered}$

Factor the difference of two cubes:

$\begin{gathered} x^3-3^3=(x-3)(x^2+3x+3^2) \\ so\colon \\ x^2(x-3)(x^2+3x+9) \\ \end{gathered}$

Therefore, the real roots are:

$\begin{gathered} x=0 \\ or \\ x=3 \end{gathered}$

Where the root:

$x=0$

has a multiplicity of 2