# Factoring by Grouping and Regrouping

Factoring by Grouping and Regrouping (factoring in "pairs") is a method for factoring polynomials, that can be applied sometimes, when terms don't have common factors.

Note, that this technique is not always applicable, because there are cases, when it is very hard (or even impossible) to see possibilities for factoring.

Idea of factoring by grouping is following: group terms and apply multiple times factoring common factor method.

Example 1. ${x}{y}-{2}{x}+{3}{y}-{6}$.

Terms don't have common factors, but if we split terms, then we can factor them separately.

${x}{y}-{2}{x}+{3}{y}-{6}=$

$={\color{red}{{{x}{y}-{2}{x}}}}{\color{green}{{+{3}{y}-{6}}}}=$ (split terms)

$={\color{red}{{{x}}}}{\color{blue}{{{\underline{{{\color{red}{{{\left({y}-{2}\right)}}}}}}}}}}{\color{green}{{+{3}}}}{\color{blue}{{{\underline{{{\color{green}{{{\left({y}-{2}\right)}}}}}}}}}}=$ (factor)

$={\left({x}+{3}\right)}{\left({y}-{2}\right)}$ (factor once more)

Answer: ${x}{y}-{2}{x}+{3}{y}-{6}={\left({x}+{3}\right)}{\left({y}-{2}\right)}$.

Sometimes, we need to regroup terms for method to work.

Example 2. Factor the following: ${5}{y}-{2}{x}{y}-{10}{x}+{{y}}^{{2}}$.

Let's try to split terms:

${5}{y}-{2}{x}{y}-{10}{x}+{{y}}^{{2}}={\color{red}{{{5}{y}-{2}{x}{y}}}}{\color{green}{{-{10}{x}+{{y}}^{{2}}}}}={y}{\left({5}-{2}{x}\right)}+{\left(-{10}{x}+{{y}}^{{2}}\right)}$.

As can be seen, above two terms have no common factors.

Thus, we can't proceed further.

However, we can regroup terms and try again.

${5}{y}-{2}{x}{y}-{10}{x}{\color{ma\genta}{{+{{y}}^{{2}}}}}=$

$={5}{y}{\color{ma\genta}{{+{{y}}^{{2}}}}}-{2}{x}{y}-{10}{x}=$ (regroup)

$={\color{red}{{{5}{y}+{{y}}^{{2}}}}}{\color{green}{{-{2}{x}{y}-{10}{x}}}}=$ (group)

$={y}{\left({5}+{y}\right)}-{2}{x}{\left({y}+{5}\right)}=$ (factor)

$={y}{\underline{{{\left({y}+{5}\right)}}}}-{2}{x}{\underline{{{\left({y}+{5}\right)}}}}=$ (according to commutative property of addition, ${5}+{y}={y}+{5}$)

$={\left({y}-{2}{x}\right)}{\left({y}+{5}\right)}$ (factor again)

Answer: ${5}{y}-{2}{x}{y}-{10}{x}+{{y}}^{{2}}={\left({y}-{2}{x}\right)}{\left({y}+{5}\right)}$.

Of course, we can split the expression into more that two terms.

Example 3. Factor ${{x}}^{{3}}+{{x}}^{{2}}{y}-{2}{{x}}^{{2}}-{2}{x}{y}-{x}-{y}$.

This example can be solved in two ways. Each way is a just matter of grouping and regrouping.

First way.

${{x}}^{{3}}+{{x}}^{{2}}{y}-{2}{{x}}^{{2}}-{2}{x}{y}-{x}-{y}=$

$={\color{red}{{{{x}}^{{3}}+{{x}}^{{2}}{y}}}}{\color{green}{{-{2}{{x}}^{{2}}-{2}{x}{y}}}}{\color{blue}{{-{x}-{y}}}}=$ (group)

$={{x}}^{{2}}{\underline{{{\left({x}+{y}\right)}}}}-{2}{x}{\underline{{{\left({x}+{y}\right)}}}}-{1}{\underline{{{\left({x}+{y}\right)}}}}=$ (factor)

$={\left({x}+{y}\right)}{\left({{x}}^{{2}}-{2}{x}-{1}\right)}$ (factor once more)

Second way.

${{x}}^{{3}}+{{x}}^{{2}}{y}-{2}{{x}}^{{2}}-{2}{x}{y}-{x}-{y}=$

$={{x}}^{{2}}{y}-{2}{x}{y}-{y}+{{x}}^{{3}}-{2}{{x}}^{{2}}-{x}=$ (regroup)

$={\color{red}{{{{x}}^{{2}}{y}-{2}{x}{y}-{y}}}}{\color{green}{{+{{x}}^{{3}}-{2}{{x}}^{{2}}-{x}}}}=$ (group)

$={y}{\left({{x}}^{{2}}-{2}{x}-{1}\right)}+{x}{\left({{x}}^{{2}}-{2}{x}-{1}\right)}=$ (factor)

$={\left({{x}}^{{2}}-{2}{x}-{1}\right)}{\left({y}+{x}\right)}$ (factor once more)

Both ways gave the same answer.

Answer: ${{x}}^{{3}}+{{x}}^{{2}}-{2}{{x}}^{{2}}-{2}{x}{y}-{x}-{y}={\left({{x}}^{{2}}-{2}{x}-{1}\right)}{\left({y}+{x}\right)}$.

Sometimes, we need to rewrite (split) term in order to factor by grouping.

Example 4. Factor the following: ${{x}}^{{2}}+{2}{x}-{3}$.

We have odd number of terms here, so can't apply grouping method.

However, we can rewrite ${2}{x}$ as ${3}{x}-{x}$.

${{x}}^{{2}}{\color{red}{{+{2}{x}}}}-{3}=$

$={{x}}^{{2}}{\color{red}{{+{3}{x}-{x}}}}-{3}=$ (rewrite the term)

$={\color{green}{{{{x}}^{{2}}+{3}{x}}}}{\color{brown}{{-{x}-{3}}}}=$ (group)

$={x}{\left({x}+{3}\right)}-{1}{\left({x}+{3}\right)}=$ (factor)

$={\left({x}+{3}\right)}{\left({x}-{1}\right)}$ (factor once more)

Answer: ${{x}}^{{2}}+{2}{x}-{3}={\left({x}+{3}\right)}{\left({x}-{1}\right)}$.

Now, it is time to exercise.

Exercise 1. Factor ${3}{{y}}^{{2}}-{6}{x}{y}-{5}{y}+{10}{x}$.

Answer: ${\left({y}-{2}{x}\right)}{\left({3}{y}-{5}\right)}$.

Exercise 2. Factor the following ${10}{{a}}^{{2}}-{12}{b}+{8}{a}-{15}{a}{b}$.

Answer: ${\left({2}{a}-{3}{b}\right)}{\left({5}{a}+{4}\right)}$. Hint: regroup.

Exercise 3. Factor ${2}{a}{{b}}^{{2}}-{5}{a}{b}+{7}{a}-{6}{{b}}^{{3}}+{15}{{b}}^{{2}}-{21}{b}$ without and with regrouping.

Answer: ${\left({a}-{3}{b}\right)}{\left({2}{{b}}^{{2}}-{5}{b}+{7}\right)}$.

Exercise 4. Factor the following: ${{y}}^{{2}}-{35}-{2}{y}$.

Answer: ${\left({y}-{7}\right)}{\left({y}+{5}\right)}$. Hint: ${{y}}^{{2}}-{35}-{2}{y}={{y}}^{{2}}-{2}{y}-{35}={{y}}^{{2}}-{7}{y}+{5}{y}-{35}$.

Exercise 5. Factor ${{x}}^{{2}}+{3}{x}{y}-{10}{{y}}^{{2}}$.

Answer: ${\left({x}+{5}{y}\right)}{\left({x}-{2}{y}\right)}$. Hint: ${3}{x}{y}=-{2}{x}{y}+{5}{x}{y}$, so ${{x}}^{{2}}+{3}{x}{y}-{10}{{y}}^{{2}}={\color{red}{{{{x}}^{{2}}-{2}{x}{y}}}}{\color{green}{{+{5}{x}{y}-{10}{{y}}^{{2}}}}}$.