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.

\begin{array}{ccc} a^{2} + 2 a b + b^{2} & = & {(a + b)}^{2} \\ and \\ a^{2} - 2 a b + b^{2} & = & {(a - b)}^{2} \end{array}

$\begin{array}{ccc}\hfill {a}^{2}+2ab+{b}^{2}& =& {\left(a+b\right)}^{2}\hfill \\ & \text{and}& \\ \hfill {a}^{2}-2ab+{b}^{2}& =& {\left(a-b\right)}^{2}\hfill \end{array}$

$\\$

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:

a^{2} + 2 a b + b^{2} = {(a + b)}^{2}

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

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

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

Example: Factoring a Perfect Square Trinomial

Factor $25{x}^{2}+20x+4$ .

Show Solution

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

Try It

Factor $49{x}^{2}-14x+1$ .

Show Solution

${\left(7x - 1\right)}^{2}$

Factoring a Difference of Squares

A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.

a^{2} - b^{2} = (a + b) (a - b)

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

$\\$

We can use this equation to factor any differences of squares.

A General Note: Differences of Squares

A difference of squares can be rewritten as two factors containing the same terms but opposite signs.

a^{2} - b^{2} = (a + b) (a - b)

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

How To: Given a difference of squares, factor it into binomials

Confirm that the first and last term are perfect squares.
Write the factored form as $\left(a+b\right)\left(a-b\right)$ .

Example: Factoring a Difference of Squares

Factor $9{x}^{2}-25$ .

Show Solution

Notice that $9{x}^{2}$ and $25$ are perfect squares because $9{x}^{2}={\left(3x\right)}^{2}$ and $25={5}^{2}$ . The polynomial represents a difference of squares and can be rewritten as $\left(3x+5\right)\left(3x - 5\right)$ .

Try It

Factor $81{y}^{2}-100$ .

Show Solution

$\left(9y+10\right)\left(9y - 10\right)$

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.

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.

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

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

$\\$

Similarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

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

$\\$

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.

x^{3} - 2^{3} = (x - 2) (x^{2} + 2 x + 4)

${x}^{3}-{2}^{3}=\left(x - 2\right)\left({x}^{2}+2x+4\right)$

The sign of the first 2 is the same as the sign between ${x}^{3}-{2}^{3}$ . The sign of the $2x$ term is opposite the sign between ${x}^{3}-{2}^{3}$ . 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

a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})

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

We can factor the difference of two cubes as

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})

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

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

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

Example: Factoring a Sum of Cubes

Factor ${x}^{3}+512$ .

Show Solution

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

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 $216{a}^{3}+{b}^{3}$ .

Show Solution

$\left(6a+b\right)\left(36{a}^{2}-6ab+{b}^{2}\right)$

Example: Factoring a Difference of Cubes

Factor $8{x}^{3}-125$ .

Show Solution

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

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: $1,000{x}^{3}-1$ .

Show Solution

$\left(10x - 1\right)\left(100{x}^{2}+10x+1\right)$

In the following two video examples we show more binomials that can be factored as a sum or difference of cubes.

Factoring Expressions with Fractional or Negative Exponents

Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, $2{x}^{\frac{1}{4}}+5{x}^{\frac{3}{4}}$ can be factored by pulling out ${x}^{\frac{1}{4}}$ and being rewritten as ${x}^{\frac{1}{4}}\left(2+5{x}^{\frac{1}{2}}\right)$ .

Example: Factoring an Expression with Fractional or Negative Exponents

Factor $3x{\left(x+2\right)}^{\frac{-1}{3}}+4{\left(x+2\right)}^{\frac{2}{3}}$ .

Show Solution

Factor out the term with the lowest value of the exponent. In this case, that would be ${\left(x+2\right)}^{-\frac{1}{3}}$ .

\begin{array}{cc} {(x + 2)}^{- \frac{1}{3}} (3 x + 4 (x + 2)) & Factor out the GCF . \\ {(x + 2)}^{- \frac{1}{3}} (3 x + 4 x + 8) & Simplify . \\ {(x + 2)}^{- \frac{1}{3}} (7 x + 8) \end{array}

$\begin{array}{cc}{\left(x+2\right)}^{-\frac{1}{3}}\left(3x+4\left(x+2\right)\right)\hfill & \text{Factor out the GCF}.\hfill \\ {\left(x+2\right)}^{-\frac{1}{3}}\left(3x+4x+8\right)\hfill & \text{Simplify}.\hfill \\ {\left(x+2\right)}^{-\frac{1}{3}}\left(7x+8\right)\hfill & \end{array}$

The following video provides more examples of how to factor expressions with negative exponents.

Try It

Factor $2{\left(5a - 1\right)}^{\frac{3}{4}}+7a{\left(5a - 1\right)}^{-\frac{1}{4}}$ .

Show Solution

${\left(5a - 1\right)}^{-\frac{1}{4}}\left(17a - 2\right)$

The following video provides more examples of how to factor expressions with fractional exponents.

Course Prerequisites