Tutorial: Factoring algebraic expressions

Tutorial: Multiplying and factoring algebraic expressions

Go to Part A: Multiplying algebraic expressions

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)

%%Note You need to understand how to multiply algebraic expressions using the distributive law before starting work on this tutorial. If you feel you need to review that material, go back to Part A by clicking on its link above.

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:

$x^{-1/2} = \frac{1}{x^{1/2}},$ and so, as we consider only expressions that do not involve fractions here—even in disguise using negative exponents—the answer is "no."

Common factors

We can think of factoring as applying the distributive law in reverse. For example,

$2x^2 + x = x(2x + 1),$ which you can check by applying the distributive law to the right-hand side. The first technique of factoring we look at is to locate a common factor: a term that occurs as a factor in each of the expressions being added or subtracted. For example, $x$ is a common factor in $2x^2 + x,$ since it is a factor of both $2x^2$ and $x.$ On the other hand, $x^2$ is not a common factor, since it is not a factor of the second term, $x.$

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 $a:$

$\color{#c1026f}{a}\color{#026fc1}{b} \pm \color{#c1026f}{a}\color{#0ea05e}{c} = \color{#c1026f}{a}(\color{#026fc1}{b} \pm \color{#0ea05e}{c}).$ Suggested video for this topic: Video by MrPetersonMaths

%%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 $2x^2y+6xy^2-6x^2y^2$, #[we can factor it out][podemos sacarlo afuera]#:

$2x^2y+xy^2-x^2y^2 \ = \ x(2xy+6y^2-6xy^2)$

#[However, $2xy$ is another common factor in $2x^2y+6xy^2-6x^2y^2$; it is the greatest common factor:

Its coefficient is positive.
It cannot be multiplied by anything except $-1$ and still remain a factor.

So, we can also write][Sin embargo, $xy$ es un otro factor común en $2x^2y+6xy^2-6x^2y^2$; es el factor común mayor:

Su coeficiente es positivo.
No se puede multiplicar por nada excepto $-1$ y todavía seguir siendo un factor.

Así, podemos también escribir]#

$2x^2y+6xy^2-6x^2y^2 \ = \ 2xy(x+3y-3xy).$ \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.