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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 2x3−8x{2}{{x}}^{{3}}-{8}{x} completely.

2x3−8x={2}{{x}}^{{3}}-{8}{x}=

=2x(x2−4)=={2}{x}{\left({{x}}^{{2}}-{4}\right)}= (factor out 2x{2}{x})

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

Answer: 2x3−8x=2x(x−2)(x+2){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: −y4−y2+2-{{y}}^{{4}}-{{y}}^{{2}}+{2}.

−y4−y2+2=-{{y}}^{{4}}-{{y}}^{{2}}+{2}=

=−(y4+y2−2)==-{\left({{y}}^{{4}}+{{y}}^{{2}}-{2}\right)}= (factor out −1-{1})

=−(y2+2)(y2−1)==-{\left({{y}}^{{2}}+{2}\right)}{\left({{y}}^{{2}}-{1}\right)}= (factor quadratics)

=−(y2+2)(y−1)(y+1)==-{\left({{y}}^{{2}}+{2}\right)}{\left({y}-{1}\right)}{\left({y}+{1}\right)}= (apply difference of squares formula)

Answer: −y4−y2+2=−(y2+2)(y−1)(y+1)-{{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 x12−1{{x}}^{{12}}-{1} completely.

x12−1=(x4)3−13={{x}}^{{12}}-{1}={{\left({{x}}^{{4}}\right)}}^{{3}}-{{1}}^{{3}}=

=(x4−1)(x8+x4+1)=={\left({{x}}^{{4}}-{1}\right)}{\left({{x}}^{{8}}+{{x}}^{{4}}+{1}\right)}= (difference of cubes)

=(x2−1)(x2+1)(x8+x4+1)=={\left({{x}}^{{2}}-{1}\right)}{\left({{x}}^{{2}}+{1}\right)}{\left({{x}}^{{8}}+{{x}}^{{4}}+{1}\right)}= (difference of squares)

=(x−1)(x+1)(x2+1)(x8+x4+1)={\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: x12−1=(x−1)(x+1)(x2+1)(x8+x4+1){{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 8x8+125x2{8}{{x}}^{{8}}+{125}{{x}}^{{2}} completely.

Answer: x2(2x2+5)(4x4−10x2+25){{x}}^{{2}}{\left({2}{{x}}^{{2}}+{5}\right)}{\left({4}{{x}}^{{4}}-{10}{{x}}^{{2}}+{25}\right)}.

Exercise 2. Factor completely: −2y3−16y2−30y-{2}{{y}}^{{3}}-{16}{{y}}^{{2}}-{30}{y}.

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

Exercise 3. Factor 512x9−y9{512}{{x}}^{{9}}-{{y}}^{{9}} completely.

Answer: (2x−y)(4x2+2xy+y2)(64x6+8x3y3+y6){\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)}.