The Big Idea: Factoring is unmultiplying.
Example $$8 x t^{3}-20 x^{4} t^{2}=4xt^{2}(2t-5x^{3})$$
Finding the GCF
The GCF of a polynomial is the collection of all common factors of every term.
Example: Find the GCF of $8 x t^{3}-20 x^{4} t^{2}$.
First we list the factors of each term and highlight common factors.
$8 x t^{3} =\color{magenta}{2} \cdot \color{magenta}{2} \cdot 2 \cdot \color{magenta}{x} \cdot \color{magenta}{t} \cdot \color{magenta}{t} \cdot t =\color{magenta}{4xt^2}\cdot 2t $
$20x^4t^2 =\color{magenta}{2} \cdot \color{magenta}{2} \cdot 5 \cdot \color{magenta}{x} \cdot x \cdot x \cdot x \cdot \color{magenta}{t} \cdot \color{magenta}{t} =\color{magenta}{4xt^2}\cdot 5x^3 $
GCF: $\color{magenta}{4xt^2}$
Notice that the GCF takes the lowest power of each factor.
$8 x t^{3} =\color{magenta}{2} \cdot \color{magenta}{2} \cdot 2 \cdot \color{magenta}{x} \cdot \color{magenta}{t} \cdot \color{magenta}{t} \cdot t =\color{magenta}{4xt^2}\cdot 2t $
$20x^4t^2 =\color{magenta}{2} \cdot \color{magenta}{2} \cdot 5 \cdot \color{magenta}{x} \cdot x \cdot x \cdot x \cdot \color{magenta}{t} \cdot \color{magenta}{t} =\color{magenta}{4xt^2}\cdot 5x^3 $
GCF: $\color{magenta}{4xt^2}$
Notice that the GCF takes the lowest power of each factor.
Examples: Find the GCF of each polynomial and factor it out.
$8 x t^{3}-20 x^{4} t^{2}$
$$
\begin{array}{lll}
\displaystyle 8 x t^{3}-20 x^{4} t^{2} &\displaystyle=\color{magenta}{4xt^2}\cdot 2t-\color{magenta}{4xt^2}\cdot 5x^3 &\mbox{}\\
\displaystyle &\displaystyle=\color{magenta}{4xt^2}(2t-5x^3) &\mbox{undistribute (factor out GCF)}\\
\end{array}
$$
$6 g^{2}+12 g^{3}+10 g$
$6g^2= \color{magenta}{2} \cdot 3 \cdot \color{magenta}{g} \cdot g$
$12g^3=\color{magenta}{2} \cdot 2 \cdot 3 \cdot \color{magenta}{g} \cdot g \cdot g$
$10g=\color{magenta}{2} \cdot 5 \cdot \color{magenta}{g}$
GCF: $\color{magenta}{2g}$
The most we can factor out of $6 g^{2}+12 g^{3}+10 g$ is a factor $2$ and a factor of $g.$ $$ \begin{array}{lll} \displaystyle 6 g^{2}+12 g^{3}+10 g&\displaystyle=\color{magenta}{2g}\cdot 3g+\color{magenta}{2g}\cdot 6g^2 +\color{magenta}{2g}\cdot 5 &\mbox{}\\ \displaystyle &\displaystyle=\color{magenta}{2g}(3g+6g^2+5) &\mbox{undistribute}\\ \end{array} $$
$12g^3=\color{magenta}{2} \cdot 2 \cdot 3 \cdot \color{magenta}{g} \cdot g \cdot g$
$10g=\color{magenta}{2} \cdot 5 \cdot \color{magenta}{g}$
GCF: $\color{magenta}{2g}$
The most we can factor out of $6 g^{2}+12 g^{3}+10 g$ is a factor $2$ and a factor of $g.$ $$ \begin{array}{lll} \displaystyle 6 g^{2}+12 g^{3}+10 g&\displaystyle=\color{magenta}{2g}\cdot 3g+\color{magenta}{2g}\cdot 6g^2 +\color{magenta}{2g}\cdot 5 &\mbox{}\\ \displaystyle &\displaystyle=\color{magenta}{2g}(3g+6g^2+5) &\mbox{undistribute}\\ \end{array} $$
$12a^3 b - 18a^2 b^2 + 30ab^3$
We'll find the GCF this time by finding the lowest power on each factor in all terms.
GCF: $2\cdot 3 \cdot a \cdot b=6ab$
Then $$ \begin{array}{lll} \displaystyle 12a^3 b - 18a^2 b^2 + 30ab^3&\displaystyle=6ab\cdot 2a^2-6ab\cdot 3ab+6ab \cdot 5b^2 &\mbox{}\\ \displaystyle &\displaystyle=6ab(2a^2-3ab+5b^2) &\mbox{}\\ \end{array} $$
GCF: $2\cdot 3 \cdot a \cdot b=6ab$
Then $$ \begin{array}{lll} \displaystyle 12a^3 b - 18a^2 b^2 + 30ab^3&\displaystyle=6ab\cdot 2a^2-6ab\cdot 3ab+6ab \cdot 5b^2 &\mbox{}\\ \displaystyle &\displaystyle=6ab(2a^2-3ab+5b^2) &\mbox{}\\ \end{array} $$
$(12 f-8) (16 a) + (12 f-8) (20 x)$
Sometimes a s GCF can contain multiple terms.
Here the GCF is $\color{magenta}{12f-8}.$
Then $$ \begin{array}{lll} \displaystyle (\color{magenta}{12f-8}) (16 a) + (\color{magenta}{12f-8}) (20 x)&\displaystyle=(\color{magenta}{12f-8})(16a+20x) &\mbox{}\\ \end{array} $$
Here the GCF is $\color{magenta}{12f-8}.$
Then $$ \begin{array}{lll} \displaystyle (\color{magenta}{12f-8}) (16 a) + (\color{magenta}{12f-8}) (20 x)&\displaystyle=(\color{magenta}{12f-8})(16a+20x) &\mbox{}\\ \end{array} $$
Factoring By Grouping: Polynomials with $4$ terms are often factored by grouping.
Examples
$35 c s+21 s+10 c r+6 r$
$$
\begin{array}{lll}
\displaystyle \overbrace{35 c s+21 s}^{\mbox{Group}}+\overbrace{10 c r+6 r}^\mbox{Group}&\displaystyle=7s(5c+3)+2r(5c+3) &\mbox{factor our GCF from grouped terms}\\
\displaystyle &\displaystyle=(7s+2r)(5c+3) &\mbox{factor out the new common factor of $5c+3$}\\
\end{array}
$$
$21 u w-35 w+18 u t-30 t$
$$
\begin{array}{lll}
\displaystyle 21 u w-35 w+18 u t-30 t&\displaystyle=7w(3u-5)+6t(3u-5) &\mbox{factor our GCF from grouped terms}\\
\displaystyle &\displaystyle=(7w+6t)(3u-5) &\mbox{factor out the new common factor of $3u-5$}\\
\end{array}
$$
Question: Will any grouping lead us to the correct factorization?
Grumpy Cat Answer:
Example : We may need to rearrange some terms before we can apply factoring by grouping.
$6 a b-35 v y+15 b y-14 a v$
First, we'll rearrange terms so that our grouping has at least some common factors.
$$
\begin{array}{lll}
\displaystyle 6 a b-35 v y+15 b y-14 a v&\displaystyle=6 a b+15 b y-35 v y-14 a v &\mbox{rearrange terms}\\
\displaystyle &\displaystyle=3b(2 a+5 y)-7v(5y+2 a) &\mbox{factor out minus!}\\
\displaystyle &\displaystyle=3b(2 a+5 y)-7v(2a+5y) &\mbox{reorder to see common factor $2a+5y$}\\
\displaystyle &\displaystyle=(3b-7v)(2a+5y) &\mbox{factor out $2a+5y$}\\
\end{array}
$$