
## Section3.6Factoring Binomials

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*}

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*}
###### Example3.6.1

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

Solution

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 Table 3.6.2 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*}
 $a^3=8x^3$ $b^3=27$ $a=2x$ $b=3$ $a^2=4x^2$ $b^2=9$ $ab=6x$
###### Example3.6.3

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

Solution

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*}
 $a^3=1$ $b^3=64x^{15}$ $a=1$ $b=4x^5$ $a^2=1$ $b^2=16x^{10}$ $ab=4x^5$

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

### Subsection3.6.1Exercises

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 Table 3.6.5

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

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

 $a^2=x^{10}$ $b^2=25y^4$ $a=x^5$ $b=5y^2$ $ab=5x^5y^2$
###### 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 Table 3.6.7

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

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

 $a^3=8x^3$ $b^3=y^6$ $a=2x$ $b=y^2$ $a^2=4x^2$ $b^2=y^4$ $ab=2xy^2$
###### 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 Table 3.6.9

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

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

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

###### 4

$36p^2-q^2$

Solution

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

###### 5

$36p^2+q^2$

Solution

$36p^2+q^2$ is prime.

###### 6

$125+y^3$

Solution

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

###### 7

$8x^3-27y^3$

Solution

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