CCSS Math: HSA.APR.D.6
Have you learned the basics of rational expression simplification? Great! Now gain more experience with some trickier examples.

### What you should be familiar with before taking this lesson

A rational expression is a ratio of two polynomials. A rational expression is considered simplified if the numerator and denominator have no factors in common.
If this is new to you, we recommend that you check out our intro to simplifying rational expressions.

### What you will learn in this lesson

In this lesson, you will practice simplifying more complicated rational expressions. Let's look at two examples, and then you can try some problems!

## Example 1: Simplifying $~\dfrac{10x^3}{2x^2-18x}$

Step 1: Factor the numerator and denominator
Here it is important to notice that while the numerator is a monomial, we can factor this as well.
$\dfrac{10x^3}{2x^2-18x}=\dfrac{ 2\cdot 5\cdot x\cdot x^2}{ 2\cdot x\cdot (x-9)}$
Step 2: List restricted values
From the factored form, we see that ${x\neq0}$ and ${x\neq9}$.
Step 3: Cancel common factors
\begin{aligned}\dfrac{ \tealD 2\cdot 5\cdot \purpleC{x}\cdot x^2}{ \tealD 2\cdot \purpleC{x}\cdot (x-9)}&=\dfrac{ \tealD{\cancel{ 2}}\cdot 5\cdot \purpleC{\cancel{x}}\cdot x^2}{ \tealD{\cancel{ 2}}\cdot \purpleC{\cancel{x}}\cdot (x-9)}\\ \\ &=\dfrac{5x^2}{x-9} \end{aligned}
We write the simplified form as follows:
$\dfrac{5x^2}{x-9}$ for $x\neq 0$
Recall that the original expression requires $x\neq 0,9$. The simplified expression must have the same undefined values.
Because of this, we must note that $x\neq 0$. We do not need to note that $x\neq 9$, since this is understood from the expression.

### Main takeaway

In this example, we see that sometimes we will have to factor monomials in order to simplify a rational expression.

1) Simplify $\dfrac{6x^2}{12x^4-9x^3}$.
Step 1: Factor the numerator and denominator
$\dfrac{6x^2}{12x^4-9x^3}=\dfrac{ 3\cdot 2\cdot x^2}{ 3x^3(4x-3)}$
Step 2: List restricted values
From the factored form, we see that ${x\neq0}$ and ${x\neq\dfrac34}$.
Step 3: Cancel common factors
\begin{aligned}\dfrac{3\cdot 2 \cdot x^2}{ 3x^3(4x-3)}&=\dfrac{ \purpleC{3}\cdot 2\cdot \tealD{x^2}}{ \purpleC{3}\cdot x\cdot \tealD{x^2}(4x-3)}\\ \\ &=\dfrac{ \purpleC{\cancel{3}}\cdot 2\cdot \tealD{\cancel{x^2}}}{ \purpleC{\cancel{3}}\cdot x\cdot \tealD{\cancel{x^2}}(4x-3)}\\ \\ &=\dfrac{2}{x(4x-3)} \end{aligned}
$\dfrac{2}{x(4x-3)}$
Notice that the original requires $x\neq 0,\dfrac34$. The simplified expression has the same requirements. We don't need to restrict any additional values.

## Example 2: Simplifying $~\dfrac{(3-x)(x-1)}{(x-3)(x+1)}$

Step 1: Factor the numerator and denominator
While it does not appear that there are any common factors, $x-3$ and $3-x$ are related. In fact, we can factor $-1$ out of the numerator to reveal a common factor of $x-3$.
\begin{aligned} &\phantom{=}\dfrac{(3-x)(x-1)}{(x-3)(x+1)} \\\\ &=\dfrac{\goldD{-1}{(-3+x)}(x-1)}{{(x-3)}(x+1)} \\\\ &=\dfrac{{-1}{\tealD{(x-3)}}(x-1)}{{\tealD{(x-3)}}(x+1)}\quad\small{\gray{\text{Commutativity}}} \end{aligned}
Step 2: List restricted values
From the factored form, we see that ${x\neq3}$ and ${x\neq-1}$.
Step 3: Cancel common factors
\begin{aligned} &\phantom{=}\dfrac{{-1}{\tealD{(x-3)}}(x-1)}{{\tealD{(x-3)}}(x+1)}\\\\\\ &=\dfrac{{-1}{\tealD{\cancel{(x-3)}}}(x-1)}{{\tealD{\cancel{(x-3)}}}(x+1)} \\\\ &=\dfrac{-1(x-1)}{x+1} \\\\ &=\dfrac{1-x}{x+1} \end{aligned}
The last step of multiplying the $-1$ into the numerator wasn't necessary, but it is common to do so.
We write the simplified form as follows:
$\dfrac{1-x}{x+1}$ for $x\neq 3$
Recall that the original expression requires $x\neq 3,-1$. The simplified expression must have the same undefined values.
Because of this, we must note that $x\neq 3$. We do not need to note that $x\neq -1$, since this is understood from the expression.

### Main takeaway

The factors $x-3$ and $3-x$ are opposites since $-1\cdot (x-3)=3-x$.
In this example, we saw that these factors canceled, but that a factor of $-1$ was added. In other words, the factors $x-3$ and $3-x$ canceled to $\textit{-1}$.
In general opposite factors $a-b$ and $b-a$ will cancel to $-1$ provided that $a\neq b$.

2) Simplify $\dfrac{(x-2)(x-5)}{(2-x)(x+5)}$.
Step 1: Factor the numerator and denominator
The numerator and denominator are already in factored form.
$\dfrac{(x-2)(x-5)}{(2-x)(x+5)}$
Step 2: List restricted values
From the factored form, we see that ${x\neq2}$ and ${x\neq-5}$.
Step 3: Cancel opposite factors
We can cancel $\tealD{(x-2)}$ and $\tealD{(2-x)}$ to a factor of $\goldD{-1}$.
\begin{aligned}\dfrac{\tealD{(x-2)}(x-5)}{\tealD{(2-x)}(x+5)}&=\dfrac{\goldD{-1}\tealD{\cancel{(x-2)}}(x-5)}{\tealD{\cancel{(2-x)}}(x+5)}\\ \\ &=\dfrac{-1(x-5)}{x+5}\\ \\ &=\dfrac{5-x}{x+5} \end{aligned}
We write the simplified form as follows:
$\dfrac{5-x}{x+5}$ for $x\neq 2$
3) Simplify $\dfrac{15-10x}{8x^3-12x^2}$.
for $x\neq$
Step 1: Factor the numerator and denominator
$\dfrac{15-10x}{8x^3-12x^2}=\dfrac{ 5(3-2x)}{ 4x^2(2x-3)}$
Step 2: List restricted values
From the factored form, we see that ${x\neq0}$ and ${x\neq\dfrac32}$.
Step 3: Cancel opposite factors
We can cancel $\tealD{(3-2x)}$ and $\tealD{(2x-3)}$ to a factor of $\goldD{-1}$.
\begin{aligned}\dfrac{ 5\tealD{(3-2x)}}{ 4x^2\tealD{(2x-3)}}&=\dfrac{ 5\cdot \goldD{(-1)}\tealD{\cancel{(3-2x)}}}{ 4x^2\tealD{\cancel{(2x-3)}}}\\ \\ &=\dfrac{-5}{4x^2} \end{aligned}
We write the simplified form as follows:
$\dfrac{-5}{4x^2}$ for $x\neq \dfrac32$

## Let's try some more problems

4) Simplify $\dfrac{3x}{15x^2-6x}$.
Step 1: Factor the numerator and denominator
$\dfrac{3x}{15x^2-6x}=\dfrac{ 3 x}{ 3x(5x-2)}$
Step 2: List restricted values
From the factored form, we see that ${x\neq0}$ and ${x\neq\dfrac25}$.
Step 3: Cancel common factors
\begin{aligned}\dfrac{ \tealD{3 x}}{ \tealD{3x}(5x-2)}&=\dfrac{ \tealD{\cancel{3 x}}}{ \tealD{\cancel{3x}}(5x-2)}\\ \\ &=\dfrac{ 1}{5x-2} \end{aligned}
Note that if the numerator cancels completely, a factor of $1$ still remains.
We write the simplified form as follows:
$\dfrac{ 1}{5x-2}$ for $x\neq 0$
5) Simplify $\dfrac{3x^3-15x^2+12x}{3x-3}$.
for $x\neq$
Step 1: Factor the numerator and denominator
Notice that there is a common factor of $3x$ in the numerator. We can factor this out, then factor the resulting trinomial.
\begin{aligned}\dfrac{3x^3-15x^2+12x}{3x-3}&=\dfrac{ 3x(x^2-5x+4)}{ 3(x-1)}\\ \\ &=\dfrac{ 3x(x-4)(x-1)}{ 3(x-1)} \end{aligned}
Step 2: List restricted values
From the factored form, we see that ${x\neq1}$.
Step 3: Cancel common factors
\begin{aligned}\dfrac{ \tealD{3}\cdot x(x-4)\purpleC{(x-1)}}{ \tealD{3}\cdot \purpleC{(x-1)}}&=\dfrac{ \tealD{\cancel{3}}\cdot x(x-4)\purpleC{\cancel{(x-1)}}}{ \tealD{\cancel{3}}\cdot \purpleC{\cancel{(x-1)}}}\\ \\ &=x(x-4) \end{aligned}
We write the simplified form as follows:
$x(x-4)$ for $x\neq 1$
6) Simplify $\dfrac{6x^2-12x}{6x-3x^2}$.
Notice that there is a common factor of $6x$ in the numerator and $3x$ in the denominator.
\begin{aligned}\dfrac{6x^2-12x}{6x-3x^2}&=\dfrac{6x(x-2)}{3x(2-x)}\\ \end{aligned}
From the factored form, we see that $x\neq 0$ and $x\neq2$.
Notice that we can write $6x$ as $2\cdot 3x$. Then we can cancel $3x$ from both the numerator and the denominator.
Next we can cancel the opposite factors $(x-2)$ and $(2-x)$ to $-1$.
\begin{aligned}\phantom{\dfrac{6x(x-2)}{3x(2-x)}}&=\dfrac{2\cdot \purpleC{(x-2)}}{\purpleC{(2-x)}}\\ \\ &=\dfrac{2\cdot(\goldD{-1}) \purpleC{\cancel{(x-2)}}}{\purpleC{\cancel{(2-x)}}}\\ \\ &=-2 \end{aligned}
$-2$ for $x\neq 0,2$