Using Techniques for Factoring Together

Related calculator: Factoring Polynomials Calculator

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