Let's Learn Factoring Binomials

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Section 4.6 Factoring Binomials

Difference of Squares.

A binomial (two term polynomial) of form $a^2-b^2$ always factors into the product $(a+b)(a-b)\text{.}$ We can confirm this by applying FOIL to the expression $(a+b)(a-b)\text{.}$

$\begin{align*} (a+b)(a-b)\amp=a^2-ab+ab-b^2\\ \amp=a^2-b^2 \end{align*}$

A few simple examples follow. As always, we can check our result by expanding the factored expression.

$\begin{equation*} x^2-4=(x+2)(x-2) \end{equation*}$

$\begin{equation*} y^2-25=(y+5)(y-5) \end{equation*}$

$\begin{equation*} 36-x^2=(6+x)(6-x) \end{equation*}$

Now let's consider a few expressions that don't immediately fit the pattern. Consider $x^{10}-16\text{.}$ Hopefully we are quick to see that $16$ is the square of $4\text{.}$ To use our factor pattern successfully, we need to also recognize that $x^{10}$ is a perfect square, as is any even power of $x\text{.}$ The power-to-a-power rule of exponents relates that $(x^m)^n=x^{mn}\text{.}$ So the power of $x$ we square that results in $x^{10}$ must be half of $10\text{,}$ i.e. $5\text{.}$ Putting it all together we have:

$\begin{equation*} x^{10}-16=(x^5+4)(x^5-4) \end{equation*}$

Similar examples follow.

$\begin{equation*} y^8-9=(y^4+3)(y^4-3) \end{equation*}$

$\begin{equation*} x^{46}-1=(x^{23}+1)(x^{23}-1) \end{equation*}$

$\begin{equation*} 100-w^{12}=(10+w^6)(10-w^6) \end{equation*}$

Exploration 4.6.1.

Figure 4.6.1. Practice Factoring Difference of Squares with One Variable.

Exploration 4.6.2.

Figure 4.6.2. Practice Factoring Difference of Squares with Two Variables.

Sum and Difference of Cubes.

A binomial (two term polynomial) of form $a^3-b^3$ always factors into the product $(a-b)(a^2+ab+b^2)\text{.}$ We can confirm this by expanding the expression $(a-b)(a^2+ab+b^2)\text{.}$

$\begin{align*} (a-b)(a^2+ab+b^2)\amp=a^3+a^2b+ab^2-a^2b-ab^2-b^3\\ \amp=a^3-b^3 \end{align*}$

Similarly, a binomial of form $a^3+b^3$ always factors into the product $(a+b)(a^2-ab+b^2)\text{.}$ We can confirm this by expanding the expression $(a+b)(a^2-ab+b^2)\text{.}$

$\begin{align*} (a+b)(a^2-ab+b^2)\amp=a^3-a^2b+ab^2+a^2b-ab^2+b^3\\ \amp=a^3+b^3 \end{align*}$

Example 4.6.3.

Factor $8x^3+27$ and $8x^3-27\text{.}$

For both binomials, \(8x^3\) corresponds to what is identified in the patterns as \(a^3\) and \(27\) corresponds to what is identified in the pattern as \(b\text{.}\) The resultant expressions for \(a\text{,}\) \(b\text{,}\) \(a^2\text{,}\) and \(b^2\) and shown in Figure 4.6.4 and the factorizations are shown to the left of the table.

\begin{gather*} 8x^3+27=(2x+3)(4x^2-6x+9)\\ \\ 8x^3-27=(2x-3)(4x^2+6x+9) \end{gather*}

\(a^3=8x^3\)	\(b^3=27\)
\(a=2x\)	\(b=3\)
\(a^2=4x^2\)	\(b^2=9\)
\(ab=6x\)

Figure 4.6.4.

Example 4.6.5.

Factor the binomials $1+64x^{15}$ and $1-64x^{15}\text{.}$

Note that the power-to-a-power rule of exponents gives us \((x^5)^3=x^{(5\times 3)}\text{.}\)

\begin{gather*} 1+64x^{15}=(1+4x^5)(1-4x^5+16x^{10})\\ \\ 1-64x^{15}=(1-4x^5)(1+4x^5+16x^{10}) \end{gather*}

\(a^3=1\)	\(b^3=64x^{15}\)
\(a=1\)	\(b=4x^5\)
\(a^2=1\)	\(b^2=16x^{10}\)
\(ab=4x^5\)

Figure 4.6.6.

Exploration 4.6.3.

Figure 4.6.7. Practice Factoring Sum or Difference of Cubes with one variable.

Exploration 4.6.4.

Figure 4.6.8. Practice Factoring Sum or Difference of Cubes with two variables.

Sum of Squares.

Unless the expression also happens to be a sum of cubes, sums of squares do not factor - that is, they are prime.

$x^2+4$ is prime.

$y^4+25$ is prime.

$w^6+4x^2$ is prime.

Many folks would like $x^2+4$ to factor, so much so that they will write $x^2+4=(x+2)^2\text{.}$ Would that it were so. But alas:

$\begin{align*} (x+2)^2\amp=(x+2)(x+2)\\ \amp=x^2+2x+2x+4\\ \amp=x^2+4x+4 \end{align*}$

In summary, $x^2+4\neq (x+2)^2\text{,}$ $x^2+4x+4=(x+2)^2\text{.}$

Exercises Exercises

Factor each binomial after first completing the indicated table.

1.

Use the factor pattern $a^2-b^2=(a+b)(a-b)$ to factor $x^{10}-25y^4$ after first completing the entries in Figure 4.6.9

$a^2=$	$b^2=$
$a=$	$b=$

Figure 4.6.9.

\(x^{10}-25y^4=(x^5+5y^2)(x^5-5y^2)\)

\(a^2=x^{10}\)	\(b^2=25y^4\)
\(a=x^5\)	\(b=5y^2\)
\(ab=5x^5y^2\)

Figure 4.6.10.

2.

Use the factor pattern $a^3-b^3=(a-b)(a^2+ab+b^2)$ to factor $8x^3-y^6$ after first completing the entries in Figure 4.6.11

$a^3=$	$b^3=$
$a=$	$b=$
$a^2=$	$b^2=$
$ab=$

Figure 4.6.11.

\(8x^3-y^6=(2x-y^2)(4x^2+2xy^2+y^4)\)

\(a^3=8x^3\)	\(b^3=y^6\)
\(a=2x\)	\(b=y^2\)
\(a^2=4x^2\)	\(b^2=y^4\)
\(ab=2xy^2\)

Figure 4.6.12.

3.

Use the factor pattern $a^3+b^3=(a+b)(a^2-ab+b^2)$ to factor $125t^{12}+27x^9$ after first completing the entries in Figure 4.6.13

$a^3=$	$b^3=$
$a=$	$b=$
$a^2=$	$b^2=$
$ab=$

Figure 4.6.13.

\(a^3=125t^{12}\)	\(b^3=27x^9\)
\(a=5t^4\)	\(b=3x^3\)
\(a^2=25t^8\)	\(b^2=9x^6\)
\(ab=15t^4x^3\)

Figure 4.6.14.

\(125t^{12}+27x^9=(5t^4+3x^3)(25t^8-15t^4x^3+9x^6)\)

Factor each binomial. Check your result by expanding the factored expression. If the binomial does not factor, state that it is prime.

4.

$36p^2-q^2$

\(36p^2-q^2=(6p+q)(6p-q)\)

5.

$36p^2+q^2$

\(36p^2+q^2 \) is prime.

6.

$125+y^3$

\(125+y^3=(5+y)(25-5y+y^2)\)

7.

$8x^3-27y^3$

\(8x^3-27y^3=(2x-3y)(4x^2+6xy+9y^2)\)