# Using Techniques for Factoring Together

## Calculatrice associée: Calculatrice de polynômes de factorisation

Now, it is time to understand how to apply learned techniques together.

Recall, that we've learned following factoring techniques:

To be successful in factoring polynomials, you need to recognize when and what method to use.

Example 1. Factor ${2}{{x}}^{{3}}-{8}{x}$ completely.

${2}{{x}}^{{3}}-{8}{x}=$

$={2}{x}{\left({{x}}^{{2}}-{4}\right)}=$ (factor out ${2}{x}$)

$={2}{x}{\left({x}-{2}\right)}{\left({x}+{2}\right)}$ (apply difference of squares formula)

Answer: ${2}{{x}}^{{3}}-{8}{x}={2}{x}{\left({x}-{2}\right)}{\left({x}+{2}\right)}$.

You possibly need to perform more than two steps.

Example 2. Factor completely: $-{{y}}^{{4}}-{{y}}^{{2}}+{2}$.

$-{{y}}^{{4}}-{{y}}^{{2}}+{2}=$

$=-{\left({{y}}^{{4}}+{{y}}^{{2}}-{2}\right)}=$ (factor out $-{1}$)

$=-{\left({{y}}^{{2}}+{2}\right)}{\left({{y}}^{{2}}-{1}\right)}=$ (factor quadratics)

$=-{\left({{y}}^{{2}}+{2}\right)}{\left({y}-{1}\right)}{\left({y}+{1}\right)}=$ (apply difference of squares formula)

Answer: $-{{y}}^{{4}}-{{y}}^{{2}}+{2}=-{\left({{y}}^{{2}}+{2}\right)}{\left({y}-{1}\right)}{\left({y}+{1}\right)}$.

Let's solve one more example.

Example 3. Factor ${{x}}^{{12}}-{1}$ completely.

${{x}}^{{12}}-{1}={{\left({{x}}^{{4}}\right)}}^{{3}}-{{1}}^{{3}}=$

$={\left({{x}}^{{4}}-{1}\right)}{\left({{x}}^{{8}}+{{x}}^{{4}}+{1}\right)}=$ (difference of cubes)

$={\left({{x}}^{{2}}-{1}\right)}{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{8}}+{{x}}^{{4}}+{1}\right)}=$ (difference of squares)

$={\left({x}-{1}\right)}{\left({x}+{1}\right)}{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{8}}+{{x}}^{{4}}+{1}\right)}$ (difference of squares once more.)

Answer: ${{x}}^{{12}}-{1}={\left({x}-{1}\right)}{\left({x}+{1}\right)}{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{8}}+{{x}}^{{4}}+{1}\right)}$.

Now, it is time to exercise.

Exercise 1. Factor ${8}{{x}}^{{8}}+{125}{{x}}^{{2}}$ completely.

Answer: ${{x}}^{{2}}{\left({2}{{x}}^{{2}}+{5}\right)}{\left({4}{{x}}^{{4}}-{10}{{x}}^{{2}}+{25}\right)}$.

Exercise 2. Factor completely: $-{2}{{y}}^{{3}}-{16}{{y}}^{{2}}-{30}{y}$.

Answer: $-{2}{y}{\left({y}+{3}\right)}{\left({y}+{5}\right)}$.

Exercise 3. Factor ${512}{{x}}^{{9}}-{{y}}^{{9}}$ completely.

Answer: ${\left({2}{x}-{y}\right)}{\left({4}{{x}}^{{2}}+{2}{x}{y}+{{y}}^{{2}}\right)}{\left({64}{{x}}^{{6}}+{8}{{x}}^{{3}}{{y}}^{{3}}+{{y}}^{{6}}\right)}$.