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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*}
\(a^3=8x^3\) \(b^3=27\)
\(a=2x\) \(b=3\)
\(a^2=4x^2\) \(b^2=9\)
\(ab=6x\)
Table3.6.2
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*}
\(a^3=1\) \(b^3=64x^{15}\)
\(a=1\) \(b=4x^5\)
\(a^2=1\) \(b^2=16x^{10}\)
\(ab=4x^5\)
Table3.6.4

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

\(a^2=\) \(b^2=\)
\(a=\) \(b=\)
Table3.6.5
Solution

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

\(a^2=x^{10}\) \(b^2=25y^4\)
\(a=x^5\) \(b=5y^2\)
\(ab=5x^5y^2\)
Table3.6.6
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

\(a^3=\) \(b^3=\)
\(a=\) \(b=\)
\(a^2=\) \(b^2=\)
\(ab=\)
Table3.6.7
Solution

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

\(a^3=8x^3\) \(b^3=y^6\)
\(a=2x\) \(b=y^2\)
\(a^2=4x^2\) \(b^2=y^4\)
\(ab=2xy^2\)
Table3.6.8
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

\(a^3=\) \(b^3=\)
\(a=\) \(b=\)
\(a^2=\) \(b^2=\)
\(ab=\)
Table3.6.9
Solution
\(a^3=125t^{12}\) \(b^3=27x^9\)
\(a=5t^4\) \(b=3x^3\)
\(a^2=25t^8\) \(b^2=9x^6\)
\(ab=15t^4x^3\)

Table3.6.10

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

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

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