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Recall that when we multiplied together the two binomials $(a+b)$ and $(a-b)$ , we obtained the product ${a}^{2}-{b}^{2}$ .
$(a+b)(a-b)={a}^{2}-{b}^{2}$
Factor ${x}^{2}-16$ . Both ${x}^{2}$ and 16 are perfect squares. The terms that, when squared, produce ${x}^{2}$ and 16 are $x$ and 4, respectively. Thus,
${x}^{2}-16=(x+4)(x-4)$
We can check our factorization simply by multiplying.
$\begin{array}{lll}(x+4)(x-4)\hfill & =\hfill & {x}^{2}-4x+4x-16\hfill \\ \hfill & =\hfill & {x}^{2}-16.\hfill \end{array}$
$49{a}^{2}{b}^{4}-121$ . Both $49{a}^{2}{b}^{4}$ and 121 are perfect squares. The terms that, when squared, produce $49{a}^{2}{b}^{4}$ and 121 are $7a{b}^{2}$ and 11, respectively. Substituting these terms into the factorization form we get
$49{a}^{2}{b}^{4}-121=\mathrm{(7}a{b}^{2}+11)(7a{b}^{2}-11)$
We can check our factorization by multiplying.
$\begin{array}{lll}(7a{b}^{2}+11)(7a{b}^{2}-11)\hfill & =\hfill & 49{a}^{2}{b}^{4}-11a{b}^{2}+11a{b}^{2}-121\hfill \\ \hfill & =\hfill & 49{a}^{2}{b}^{4}-121\hfill \end{array}$
$3{x}^{2}-27$ . This doesn’t look like the difference of two squares since we don’t readily know the terms that produce $3{x}^{2}$ and 27. However, notice that 3 is common to both the terms. Factor out 3.
$3({x}^{2}-9)$
Now we see that ${x}^{2}-9$ is the difference of two squares. Factoring the ${x}^{2}-9$ we get
$\begin{array}{lll}3{x}^{2}-27\hfill & =\hfill & 3({x}^{2}-9)\hfill \\ \hfill & =\hfill & 3(x+3)(x-3)\hfill \end{array}$
Be careful not to drop the factor 3.
If possible, factor the following binomials completely.
${m}^{2}-25$
$(m+5)(m-5)$
$36{p}^{2}-81{q}^{2}$
$9(2p-3q)(2p+3q)$
$49{a}^{4}-{b}^{2}{c}^{2}$
$(7{a}^{2}+bc)(7{a}^{2}-bc)$
${x}^{8}{y}^{4}-100{w}^{12}$
$({x}^{4}{y}^{2}+10{w}^{6})({x}^{4}{y}^{2}-10{w}^{6})$
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