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Study Guides > College Algebra CoRequisite Course

Factoring Special Cases

Learning Outcomes

  • Factor a perfect square trinomial.
  • Factor a difference of squares.
  • Factor a sum and difference of cubes.
  • Factor an expression with negative or fractional exponents.

Factoring a Perfect Square Trinomial

A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.
[latex]\begin{array}{ccc}\hfill {a}^{2}+2ab+{b}^{2}& =& {\left(a+b\right)}^{2}\hfill \\ & \text{and}& \\ \hfill {a}^{2}-2ab+{b}^{2}& =& {\left(a-b\right)}^{2}\hfill \end{array}[/latex]
[latex]\\[/latex]
We can use this equation to factor any perfect square trinomial.

A General Note: Perfect Square Trinomials

A perfect square trinomial can be written as the square of a binomial:
[latex]{a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}[/latex]

How To: Given a perfect square trinomial, factor it into the square of a binomial

  1. Confirm that the first and last term are perfect squares.
  2. Confirm that the middle term is twice the product of [latex]ab[/latex].
  3. Write the factored form as [latex]{\left(a+b\right)}^{2}[/latex].

Example: Factoring a Perfect Square Trinomial

Factor [latex]25{x}^{2}+20x+4[/latex].

Answer: Notice that [latex]25{x}^{2}[/latex] and [latex]4[/latex] are perfect squares because [latex]25{x}^{2}={\left(5x\right)}^{2}[/latex] and [latex]4={2}^{2}[/latex]. Then check to see if the middle term is twice the product of [latex]5x[/latex] and [latex]2[/latex]. The middle term is, indeed, twice the product: [latex]2\left(5x\right)\left(2\right)=20x[/latex]. Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\left(5x+2\right)}^{2}[/latex].

Try It

Factor [latex]49{x}^{2}-14x+1[/latex].

Answer: [latex]{\left(7x - 1\right)}^{2}[/latex]

Q & A

Is there a formula to factor the sum of squares? No. A sum of squares cannot be factored.
Watch this video to see another example of how to factor a difference of squares. https://youtu.be/Li9IBp5HrFA

Factoring the Sum and Difference of Cubes

Now we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.
[latex]{a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)[/latex]
[latex-display]\\[/latex-display] Similarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.
[latex]{a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)[/latex]
[latex-display]\\[/latex-display] We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following example.
[latex]{x}^{3}-{2}^{3}=\left(x - 2\right)\left({x}^{2}+2x+4\right)[/latex]
The sign of the first 2 is the same as the sign between [latex]{x}^{3}-{2}^{3}[/latex]. The sign of the [latex]2x[/latex] term is opposite the sign between [latex]{x}^{3}-{2}^{3}[/latex]. And the sign of the last term, 4, is always positive.

A General Note: Sum and Difference of Cubes

We can factor the sum of two cubes as
[latex]{a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)[/latex]
We can factor the difference of two cubes as
[latex]{a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)[/latex]

How To: Given a sum of cubes or difference of cubes, factor it

  1. Confirm that the first and last term are cubes, [latex]{a}^{3}+{b}^{3}[/latex] or [latex]{a}^{3}-{b}^{3}[/latex].
  2. For a sum of cubes, write the factored form as [latex]\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)[/latex]. For a difference of cubes, write the factored form as [latex]\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)[/latex].

Example: Factoring a Sum of Cubes

Factor [latex]{x}^{3}+512[/latex].

Answer: Notice that [latex]{x}^{3}[/latex] and [latex]512[/latex] are cubes because [latex]{8}^{3}=512[/latex]. Rewrite the sum of cubes as [latex]\left(x+8\right)\left({x}^{2}-8x+64\right)[/latex].

Analysis of the Solution

After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.

Try It

Factor the sum of cubes [latex]216{a}^{3}+{b}^{3}[/latex].

Answer: [latex]\left(6a+b\right)\left(36{a}^{2}-6ab+{b}^{2}\right)[/latex]

Example: Factoring a Difference of Cubes

Factor [latex]8{x}^{3}-125[/latex].

Answer: Notice that [latex]8{x}^{3}[/latex] and [latex]125[/latex] are cubes because [latex]8{x}^{3}={\left(2x\right)}^{3}[/latex] and [latex]125={5}^{3}[/latex]. Write the difference of cubes as [latex]\left(2x - 5\right)\left(4{x}^{2}+10x+25\right)[/latex].

Analysis of the Solution

Just as with the sum of cubes, we will not be able to further factor the trinomial portion.

Try It

Factor the difference of cubes: [latex]1,000{x}^{3}-1[/latex].

Answer: [latex]\left(10x - 1\right)\left(100{x}^{2}+10x+1\right)[/latex]

The following video provides more examples of how to factor expressions with fractional exponents. https://youtu.be/R6BzjR2O4z8

Licenses & Attributions

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CC licensed content, Shared previously

  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
  • Examples: Factoring Binomials (Special). Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
  • Ex1: Factor a Sum or Difference of Cubes. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
  • Ex3: Factor a Sum or Difference of Cubes. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.
  • Question ID 7919, 7929, 7922. Authored by: Tyler Wallace. License: CC BY: Attribution. License terms: IMathAS Community License, CC-BY + GPL.
  • Question ID 93666, 93668. Authored by: Michael Jenck. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.

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