Tutorial: Multiplying and factoring algebraic equations
This tutorial: Part B: Factoring algebraic expressions: Taking out common factors
(This topic is also in Section 0.3 in Finite Mathematics and Applied Calculus)
Factors
To factor an expression is to write it as a product of other expressions, called its factors. For instance,
$12 = \color{blue}{(4)} \ \color{indianred}{(3)},$ so $\ \color{blue}{4}\ $ %%and $\ \color{indianred}{3}$ are factors of $12.$
\\ $12 = \color{blue}{(12)} \ \color{indianred}{(1)},$ so $\ \color{blue}{12}\ $ %%and $\ \color{indianred}{1}$ are also factors of $12.$ \t
\\ $2x = \color{blue}{(2)} \ \color{indianred}{(x)},$ so $\ \color{blue}{2}\ $ %%and $\ \color{indianred}{x}$ are factors of $2x.$ \t
\\ $2x^2 = \color{blue}{(2)} \ \color{indianred}{(x^2)},$ so $\ \color{blue}{2}\ $ %%and $\ \color{indianred}{x^2}$ are factors of $2x^2.$
\\ $2x^2 = \color{blue}{(2x)} \ \color{indianred}{(x)},$ so $\ \color{blue}{2x}\ $ %%and $\ \color{indianred}{x}$ are also factors of $2x^2.$
Identifying factors
%%Q: Can we say the following:
$1 = \color{blue}{(x)} \ \color{indianred}{\left(\frac{1}{x}\right)},$ so $\ \color{blue}{x}\ $ %%and $\ \color{indianred}{\frac{1}{x}}$ are factors of $1?$
%%A: The answer depends on the context. In the context of expressions that can involve fractions, the answer would be "yes," but in the rest of this tutorial we consider only expressions that do not involve fractions, so the answer is "no."
%%Q: What about this:
$x = \color{blue}{(x^{3/2})} \ \color{indianred}{x^{1/2}},$ so $\ \color{blue}{x^{3/2}}\ $ %%and $\ \color{indianred}{x^{1/2}}$ are factors of $x?$
%%A: Again, no: $x^{1/2}$ is really a fraction in disguise:  $x^{1/2} = \frac{1}{x^{1/2}},$
Common factors
We can think of factoring as applying the distributive law in reverse. For example,

$2x^2 + x = x(2x + 1),$
Taking out a common factor
Once we have located a common factor in a sum or difference, we can "factor it out" by finding the other factor in each of the summands (as we did in the first quiz of this tutorial):
In symbols: Because $a$ is a common factor in $ab \pm ac$, we can take out the common factor $a:$

$\color{#c1026f}{a}\color{#026fc1}{b} \pm \color{#c1026f}{a}\color{#0ea05e}{c} = \color{#c1026f}{a}(\color{#026fc1}{b} \pm \color{#0ea05e}{c}).$
%%Examples
Because $x$ is a common factor in $2x^2 + x$, #[we can factor it out][podemos sacarlo afuera]#:
$2x^2 + x$ \t $\ = \ \color{#c1026f}{(x)}\color{#026fc1}{(2x)} + \color{#c1026f}{(x)}\color{#0ea05e}{(1)}$
\\ \t $\ = \ \color{#c1026f}{x}(\color{#026fc1}{2x} + \color{#0ea05e}{1})$
#[Because][Ya que]# $2y^2$ is a common factor in $6y^4 + 2y^3  4y^2$, #[we can factor it out][podemos sacarlo afuera]#:
$6y^4 + 2y^3  4y^2$ \t $\ = \ \color{#c1026f}{(2y^2)}\color{#026fc1}{(y^2)} + \color{#c1026f}{(2y^2)}\color{#0ea05e}{(y)}  \color{#c1026f}{(2y^2)}\color{#a05eae}{(2)}$
\\ \t $\ = \ \color{#c1026f}{2y^2}(\color{#026fc1}{y^2} + \color{#0ea05e}{y}  \color{#a05eae}{2})$
#[In the remaining examples we will leave out the middle step (do that mentally!)][En los ejemplos que quedan vamos a omitir el paso intermedio (¡hazlo mentalmente!)]#
#[Because][Ya que]# $x$ is a common factor in $2x^2y+6xy^26x^2y^2$, #[we can factor it out][podemos sacarlo afuera]#:
$2x^2y+xy^2x^2y^2 \ = \ x(2xy+6y^26xy^2)$
#[However, $2xy$ is another common factor in $2x^2y+6xy^26x^2y^2$; it is the greatest common factor:
 Its coefficient is positive.
 It cannot be multiplied by anything except $1$ and still remain a factor.
 Su coeficiente es positivo.
 No se puede multiplicar por nada excepto $1$ y todavía seguir siendo un factor.
$2x^2y+6xy^26x^2y^2 \ = \ 2xy(x+3y3xy).$ \t \gap[40] #[Taking out the greatest common factor][Sacando el factor común mayor]#
Now try the exercises in Section 0.3 in Finite Mathematics and Applied Calculus.
or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.
Copyright © 2021 Stefan Waner and Steven R. Costenoble