The Special Forms are:
Perfect Square Trinomials
Difference of Two Squares
Sum of Cubes
Difference of Cubes
Perfect Square Trinomials: Factor the following expressions.
Recall: $(a+b)^2=a^2+2ab+b^2.$
That is, if the first and last terms of a trinomial are perfect squares, that's a hint that you might have a perfect square trinomial on your hands.
The following gives us a guide for how to use this result. $$ (\color{magenta}{\mbox{____}}+\color{blue}{\mbox{____}})^2=(\color{magenta}{\mbox{____}})^2+2(\color{blue}{\mbox{____}})(\color{magenta}{\mbox{____}})+(\color{blue}{\mbox{____}})^2 $$ Example: $25r^2-60r+36$
The first term is a perfect square: $25r^2=(\color{magenta}{5r})^2.$
The second term is a perfect square: $36=(\color{blue}{6})^2=(\color{blue}{-6})^2.$ Since the middle sign is negative, we'll try $\color{blue}{-6}.$ $$ \begin{array}{lll} \displaystyle 25r^2-60r+36&\displaystyle=(\color{magenta}{5r})^2+2(\color{magenta}{5r})(\color{blue}{-6})+(\color{blue}{-6})^2 &\mbox{}\\ %\displaystyle &\displaystyle= (\color{magenta}{5r}\color{blue}{-6})(\color{magenta}{5r}\color{blue}{-6})&\mbox{}\\ \displaystyle &\displaystyle=(\color{magenta}{5r}\color{blue}{-6})^2 &\mbox{}\\ \end{array} $$ If you check it, it works!
The second term is a perfect square: $36=(\color{blue}{6})^2=(\color{blue}{-6})^2.$ Since the middle sign is negative, we'll try $\color{blue}{-6}.$ $$ \begin{array}{lll} \displaystyle 25r^2-60r+36&\displaystyle=(\color{magenta}{5r})^2+2(\color{magenta}{5r})(\color{blue}{-6})+(\color{blue}{-6})^2 &\mbox{}\\ %\displaystyle &\displaystyle= (\color{magenta}{5r}\color{blue}{-6})(\color{magenta}{5r}\color{blue}{-6})&\mbox{}\\ \displaystyle &\displaystyle=(\color{magenta}{5r}\color{blue}{-6})^2 &\mbox{}\\ \end{array} $$ If you check it, it works!
A Difference of Two Squares
Recall: $(a+b)(a-b)=a^2-b^2.$
We use the above fact to factor expressions of the form $a^2-b^2$.
The following gives us a guide for how to use this result. $$ (\color{magenta}{\mbox{____}})^2-(\color{blue}{\mbox{____}})^2=(\color{magenta}{\mbox{____}}+\color{blue}{\mbox{____}})(\color{magenta}{\mbox{____}}-\color{blue}{\mbox{____}}) $$
Examples
$x^2-4$
$$
\begin{array}{lll}
\displaystyle x^2-4&\displaystyle=(\color{magenta}{x})^2-(\color{blue}{2})^2 &\mbox{}\\
\displaystyle &\displaystyle=(\color{magenta}{x}+\color{blue}{2})(\color{magenta}{x}-\color{blue}{2}) &\mbox{}\\
\end{array}
$$
$36x^2-16$
$$
\begin{array}{lll}
\displaystyle 36x^2-16&\displaystyle=(\color{magenta}{6x})^2-(\color{blue}{4})^2 &\mbox{}\\
\displaystyle &\displaystyle=(\color{magenta}{6x}+\color{blue}{4})(\color{magenta}{6x}-\color{blue}{4}) &\mbox{}\\
\end{array}
$$
$64s^2-y^2$
$$
\begin{array}{lll}
\displaystyle 64s^2-y^2&\displaystyle=(\color{magenta}{8s})^2-(\color{blue}{y})^2 &\mbox{}\\
\displaystyle &\displaystyle=(\color{magenta}{8s}+\color{blue}{y})(\color{magenta}{8s}-\color{blue}{y}) &\mbox{}\\
\end{array}
$$
$w^2+49p^2$
This is a sum of squares, not a difference of squares.
A sum of squares is not factorable.
A sum of squares is not factorable.
Some Vocab
Binomials which differ only in their middle signs are called conjugates.
The following are examples of conjugates. $$a+b\,\, \mbox{ and } \,\,a-b$$ $$y-2\,\, \mbox{ and } \,\,y+2$$ $$6x+4\,\, \mbox{ and } \,\,6x-4$$ $$8s-y\,\, \mbox{ and } \,\,8s+y$$ This vocab is handy for remembering that a difference of squares factors into conjugates.
Sums and Differences of Cubes
Sum of Cubes: $a^3+b^3=(a+b)(a^2-ab+b^2)$
Difference of Cubes: $a^3-b^3=(a-b)(a^2+ab+b^2)$
The following gives us a guide for how to use the above facts. $$ (\color{magenta}{\mbox{____}})^3+(\color{blue}{\mbox{____}})^3=(\color{magenta}{\mbox{____}}+\color{blue}{\mbox{____}})((\color{magenta}{\mbox{____}})^2-(\color{magenta}{\mbox{____}})(\color{blue}{\mbox{____}})+(\color{blue}{\mbox{____}})^2) $$ $$ (\color{magenta}{\mbox{____}})^3-(\color{blue}{\mbox{____}})^3=(\color{magenta}{\mbox{____}}-\color{blue}{\mbox{____}})((\color{magenta}{\mbox{____}})^2+(\color{magenta}{\mbox{____}})(\color{blue}{\mbox{____}})+(\color{blue}{\mbox{____}})^2) $$ Example: $27p^3+8$
$$
\begin{array}{lll}
\displaystyle 27p^3+8&\displaystyle=(\color{magenta}{3p})^3+(\color{blue}{2})^3 &\mbox{}\\
\displaystyle &\displaystyle=(\color{magenta}{3p}+\color{blue}{2})((\color{magenta}{3p})^2-(\color{magenta}{3p})(\color{blue}{2})+(\color{blue}{2})^2) &\mbox{}\\
\displaystyle &\displaystyle=(3p+2)(9p^2-6p+4) &\mbox{}\\
\end{array}
$$
Example: $64r^3-1$
$$
\begin{array}{lll}
\displaystyle 64r^3-1&\displaystyle=(\color{magenta}{4r})^3-(\color{blue}{1})^3 &\mbox{}\\
\displaystyle &\displaystyle=(\color{magenta}{4r}-\color{blue}{1})((\color{magenta}{4r})^2+(\color{magenta}{4r})(\color{blue}{1})+(\color{blue}{1})^2) &\mbox{}\\
\displaystyle &\displaystyle=(4r-1)(16r^2+4r+1) &\mbox{}\\
\end{array}
$$
Example: $27t^3+125u^3$
$$
\begin{array}{lll}
\displaystyle 27p^3+8&\displaystyle=(\color{magenta}{3t})^3+(\color{blue}{5u})^3 &\mbox{}\\
\displaystyle &\displaystyle=(\color{magenta}{3t}+\color{blue}{5u})((\color{magenta}{3t})^2-(\color{magenta}{3t})(\color{blue}{5u})+(\color{blue}{5u})^2) &\mbox{}\\
\displaystyle &\displaystyle=(3p+5u)(9t^2-15tu+25u^2) &\mbox{}\\
\end{array}
$$