6.5 Factoring two special products

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.

$\begin{array}{lll}(x+4)(x-4)\hfill & =\hfill & x^2-4x+4x-16\hfill \\ \hfill & =\hfill & x^2-16.\hfill \end{array}$

$49a^2{b}^4-121$ . Both $49a^2{b}^4$ and 121 are perfect squares. The terms that, when squared, produce $49a^2{b}^4$ and 121 are $7a{b}^2$ and 11, respectively. Substituting these terms into the factorization form we get

$49a^2{b}^4-121=\mathrm{(7}a{b}^2+11)(7a{b}^2-11)$

We can check our factorization by multiplying.

$\begin{array}{lll}(7a{b}^2+11)(7a{b}^2-11)\hfill & =\hfill & 49a^2{b}^4-11a{b}^2+11a{b}^2-121\hfill \\ \hfill & =\hfill & 49a^2{b}^4-121\hfill \end{array}$

$3x^2-27$ . This doesn’t look like the difference of two squares since we don’t readily know the terms that produce $3x^2$ 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

$\begin{array}{lll}3x^2-27\hfill & =\hfill & 3(x^2-9)\hfill \\ \hfill & =\hfill & 3(x+3)(x-3)\hfill \end{array}$

Be careful not to drop the factor 3.