How do you find the zeros with multiplicity for the function p(x)=(x^3-8)(x^5-4x^3)?

Nov 15, 2016

$p \left(x\right)$ has zeros:

$0$ with multiplicity $3$

$2$ with multiplicity $2$

$- 2$ with multiplicity $1$

$- 1 + \sqrt{3} i$ with multiplicity $1$

$- 1 - \sqrt{3} i$ with multiplicity $1$

Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

We find:

$p \left(x\right) = \left({x}^{3} - 8\right) \left({x}^{5} - 4 {x}^{3}\right)$

$\textcolor{w h i t e}{p \left(x\right)} = \left({x}^{3} - {2}^{3}\right) {x}^{3} \left({x}^{2} - {2}^{2}\right)$

$\textcolor{w h i t e}{p \left(x\right)} = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right) {x}^{3} \left(x - 2\right) \left(x + 2\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} {\left(x - 2\right)}^{2} \left(x + 2\right) \left({x}^{2} + 2 x + 4\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} {\left(x - 2\right)}^{2} \left(x + 2\right) \left({x}^{2} + 2 x + 1 + 3\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} {\left(x - 2\right)}^{2} \left(x + 2\right) \left({\left(x + 1\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} {\left(x - 2\right)}^{2} \left(x + 2\right) \left(\left(x + 1\right) - \sqrt{3} i\right) \left(\left(x + 1\right) + \sqrt{3} i\right)$

$\textcolor{w h i t e}{p \left(x\right)} = {x}^{3} {\left(x - 2\right)}^{2} \left(x + 2\right) \left(x + 1 - \sqrt{3} i\right) \left(x + 1 + \sqrt{3} i\right)$

Hence zeros:

$0$ with multiplicity $3$

$2$ with multiplicity $2$

$- 2$ with multiplicity $1$

$- 1 + \sqrt{3} i$ with multiplicity $1$

$- 1 - \sqrt{3} i$ with multiplicity $1$