# Factoring Common Factor

## Related calculator: Factoring Polynomials Calculator

Factoring is the process of rewriting polynomial into product of factors.

Factoring polynomials is quite similar to factoring numbers, but harder (because you need to work with variables) and not exactly the same.

Process of factoring is, in some sort, inverse process of multiplication of polynomials (FOIL etc.). You have two or more polynomials and you multiply them to get new polynomial.

Now, you have a polynomial and want to find its factors.

For example, (techniques for factoring you will learn later)

$2x+6=\color{red}{\underbrace{2}_{\text{factor}}}\color{green}{\underbrace{(x+3)}_{\text{factor}}}$

$2x^2+9x-5=\color{red}{\underbrace{(2x-1)}_{\text{factor}}}\color{green}{\underbrace{(x+5)}_{\text{factor}}}$

$v^4w-16w=\color{red}{\underbrace{w}_{\text{factor}}}\color{green}{\underbrace{(v^2+1)}_{\text{factor}}}\color{brown}{\underbrace{(v-1)}_{\text{factor}}}\color{magenta}{\underbrace{(v+1)}_{\text{factor}}}$

Factoring Common Factor

Factoring Common Factor is a technique, based on distributive property of multiplication ${a}{b}+{a}{c}={a}{\left({b}+{c}\right)}$.

We try to find common factor in all addends and then factor it out.

Example 1. Factor the following: ${3}{x}+{12}$.

We can write ${12}$ as ${3}\cdot{4}$.

${3}{x}+{12}={3}{x}+{3}\cdot{4}=$

$={\underline{{{\color{red}{{{3}}}}}}}{x}+{\underline{{{\color{red}{{{3}}}}}}}\cdot{4}=$ (common factor is 3)

$={3}{\left({x}+{4}\right)}$ (distributive property of multiplication)

Answer: ${3}{x}+{12}={3}{\left({x}+{4}\right)}$.

As can be seen, factoring is equivalent to finding numbers/expressions, that divide out each summand evenly.

Of course, variables can also be factored out.

Example 2. Factor the following: ${20}{{w}}^{{3}}{{x}}^{{2}}-{4}{{w}}^{{4}}{x}$.

Let's first write each summand, using multiplication only:

${20}{{w}}^{{3}}{{x}}^{{2}}={\color{red}{{{4}}}}\cdot{5}\cdot{\color{red}{{{w}}}}\cdot{\color{red}{{{w}}}}\cdot{\color{red}{{{w}}}}\cdot{\color{red}{{{x}}}}\cdot{x}$

$-{4}{{w}}^{{4}}{x}=-{\color{red}{{{4}}}}\cdot{\color{red}{{{w}}}}\cdot{\color{red}{{{w}}}}\cdot{\color{red}{{{w}}}}\cdot{w}\cdot{\color{red}{{{x}}}}$

I highlighted common factors, so we can factor them out:

${20}{{w}}^{{3}}{{x}}^{{2}}-{4}{{w}}^{{4}}{x}=$

$={4}\cdot{5}\cdot{w}\cdot{w}\cdot{w}\cdot{x}\cdot{x}-{4}\cdot{w}\cdot{w}\cdot{w}\cdot{w}\cdot{x}=$ (rewrite without exponents)

$={4}\cdot{w}\cdot{w}\cdot{w}\cdot{x}{\left({5}{x}-{w}\right)}=$ (factor)

$={4}{{w}}^{{3}}{x}{\left({5}{x}-{w}\right)}$ (simplify)

Be careful, with minus signs! Don't forget about them.

Answer: ${20}{{w}}^{{3}}{{x}}^{{2}}-{4}{{w}}^{{4}}{x}={4}{{w}}^{{3}}{x}{\left({5}{x}-{w}\right)}$.

Since distributive property of multiplication is valid even if there are more than two summands, then we can handle cases, when there are more than two summands.

Example 3. Factor $-{15}{{x}}^{{5}}{{y}}^{{7}}{{z}}^{{3}}-{24}{{x}}^{{3}}{{y}}^{{3}}{{z}}^{{3}}-{18}{{y}}^{{2}}{{z}}^{{10}}$.

Again, we need to write each summand as product of factors to figure out what is common.

$-{15}{{x}}^{{5}}{{y}}^{{7}}{{z}}^{{3}}={\color{red}{{{\left(-{3}\right)}}}}\cdot{5}\cdot{{x}}^{{5}}\cdot{\color{red}{{{{y}}^{{2}}}}}\cdot{{y}}^{{5}}\cdot{\color{red}{{{{z}}^{{3}}}}}$

$-{24}{{x}}^{{3}}{{y}}^{{3}}{{z}}^{{3}}={\color{red}{{{\left(-{3}\right)}}}}\cdot{8}\cdot{{x}}^{{3}}\cdot{\color{red}{{{{y}}^{{2}}}}}\cdot{y}\cdot{\color{red}{{{{z}}^{{3}}}}}$

$-{18}{{y}}^{{2}}{{z}}^{{10}}={\color{red}{{{\left(-{3}\right)}}}}\cdot{6}\cdot{\color{red}{{{{y}}^{{2}}}}}\cdot{\color{red}{{{{z}}^{{3}}}}}\cdot{{z}}^{{7}}$.

In cases of exponent, the common factor will be the variable with the lowest exponent.

Exponents of ${x}$ are 5, 3 and 0 (last term doesn't contain ${x}$), so the common factor is ${{x}}^{{0}}={1}$.

Exponents of ${y}$ are 7, 3 and 2, so the common factor is ${{y}}^{{2}}$.

Exponents of ${z}$ are 3, 3 and 10, so the common factor is ${{z}}^{{3}}$.

Now, we can perform factoring:

$-{15}{{x}}^{{5}}{{y}}^{{7}}{{z}}^{{3}}-{24}{{x}}^{{3}}{{y}}^{{3}}{{z}}^{{3}}-{18}{{y}}^{{2}}{{z}}^{{10}}=-{3}{{y}}^{{2}}{{z}}^{{3}}{\left({5}{{x}}^{{5}}{{y}}^{{5}}+{8}{{x}}^{{3}}{y}+{6}{{z}}^{{7}}\right)}$.

Again, be careful with minus sign. Here, we've factored it out, and changed signs inside parenthesis.

Answer: $-{15}{{x}}^{{5}}{{y}}^{{7}}{{z}}^{{3}}-{24}{{x}}^{{3}}{{y}}^{{3}}{{z}}^{{3}}-{18}{{y}}^{{2}}{{z}}^{{10}}=-{3}{{y}}^{{2}}{{z}}^{{3}}{\left({5}{{x}}^{{5}}{{y}}^{{5}}+{8}{{x}}^{{3}}{y}+{6}{{z}}^{{7}}\right)}$.

Last example is the case when common factor is expression/polynomial.

Example 4. Factor ${\left({x}+{1}\right)}{\left({3}{x}-{4}\right)}-{\left({x}+{1}\right)}{\left({x}-{8}\right)}$.

Common factor here is ${x}+{1}$:

${\underline{{{\color{red}{{{\left({x}+{1}\right)}}}}}}}{\left({3}{x}-{4}\right)}-{\underline{{{\color{red}{{{\left({x}+{1}\right)}}}}}}}{\left({x}-{8}\right)}=$

$={\left({x}+{1}\right)}{\left({\left({3}{x}-{4}\right)}-{\left({x}-{8}\right)}\right)}=$ (factor and be careful with parenthesis)

$={\left({x}+{1}\right)}{\left({3}{x}-{4}-{x}+{8}\right)}=$ (expand second factor)

$={\left({x}+{1}\right)}{\left({2}{x}+{4}\right)}=$ (simplify second factor)

$={2}{\left({x}+{1}\right)}{\left({x}+{2}\right)}=$ (factor 2 out of ${2}{x}+{4}$)

Answer: ${\left({x}+{1}\right)}{\left({3}{x}-{4}\right)}-{\left({x}+{1}\right)}{\left({x}-{8}\right)}={2}{\left({x}+{1}\right)}{\left({x}+{2}\right)}$.

Now, it is time to exercise.

Exercise 1. Factor the following: ${12}{x}-{18}$.

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

Exercise 2. Factor ${12}{{x}}^{{3}}+{8}{{x}}^{{2}}-{4}{x}$.

Answer: ${4}{x}{\left({3}{{x}}^{{2}}+{2}{x}-{1}\right)}$.

Exercise 3. Factor the following: $-{18}{{a}}^{{2}}{{b}}^{{2}}{{c}}^{{2}}-{12}{a}{b}{{c}}^{{2}}+{24}{{a}}^{{2}}{b}{{c}}^{{4}}$.

Answer: $-{6}{a}{b}{c}{\left({3}{a}{b}{c}+{2}{c}-{4}{a}{{c}}^{{3}}\right)}$.

Exercise 4. Factor ${\left({x}+{5}\right)}{\left({2}{x}+{1}\right)}+{\left({x}+{5}\right)}{\left({x}+{3}\right)}$.

Answer: ${\left({x}+{5}\right)}{\left({3}{x}+{4}\right)}$.

Exercise 5. Factor ${2}{\left({x}-{8}\right)}{\left({2}{x}+{3}\right)}+{2}{\left({6}{x}+{5}\right)}{\left({8}-{x}\right)}$.

Answer: $-{4}{\left({x}-{8}\right)}{\left({2}{x}+{1}\right)}$. Hint: ${8}-{x}=-{\left({x}-{8}\right)}$, so ${2}{\left({6}{x}+{5}\right)}{\left({8}-{x}\right)}=-{2}{\left({6}{x}+{5}\right)}{\left({x}-{8}\right)}$.