Intro to simplifying rational expressions

CCSS Math: HSA.APR.D.6
Learn what it means to simplify a rational expression, and how it's done!

What you should be familiar with before taking this lesson

A rational expression is a ratio of two polynomials. The domain of a rational expression is all real numbers except those that make the denominator equal to zero.
For example, the domain of the rational expression x+2x+1\dfrac{x+2}{x+1} is all real numbers except -1\textit{-1}, or x1x\neq -1.
If this is new to you, we recommend that you check out our intro to rational expressions.
You should also know how to factor polynomials for this lesson.

What you will learn in this lesson

In this article, we will learn how to simplify rational expressions by looking at several examples.

Introduction

A rational expression is considered simplified if the numerator and denominator have no factors in common.
We can simplify rational expressions in much the same way as we simplify numerical fractions.
For example, the simplified version of 68\dfrac 68 is 34\dfrac{3}{4}. Notice how we canceled a common factor of 22 from the numerator and the denominator:
68=2324Factor=2324Cancel common factors=34Simplify\begin{aligned} \dfrac68&= \dfrac{2\cdot 3}{2\cdot 4}&&\small{\gray{\text{Factor}}} \\\\ &= \dfrac{\tealD{\cancel{2}}\cdot 3}{\tealD{\cancel{2}}\cdot 4}&&\small{\gray{\text{Cancel common factors}}} \\ \\ &= \dfrac{3}{4} &&\small{\gray{\text{Simplify}}} \end{aligned}

Example 1: Simplifying x2+3xx2+5x\dfrac{x^2+3x}{x^2+5x}

Step 1: Factor the numerator and denominator
The only way to see if the numerator and denominator share common factors is to factor them!
x2+3xx2+5x=x(x+3)x(x+5)\dfrac{x^2+3x}{x^2+5x}=\dfrac{ x(x+3)}{ x(x+5)}
Step 2: List restricted values
At this point, it is helpful to notice any restrictions on xx. These will carry over to the simplified expression.
Since division by 00 is undefined, here we see that x0\blueD{x\neq0} and x5\purpleC{x\neq -5}.
x(x+3)x(x+5)\dfrac{ x(x+3)}{ \blueD x\purpleC{(x+5)}}
Step 3: Cancel common factors
Now notice that the numerator and denominator share a common factor of xx. This can be canceled out.
x(x+3)x(x+5)=x(x+3)x(x+5)=x+3x+5\begin{aligned}\dfrac{\tealD x(x+3)}{\tealD x(x+5)}&=\dfrac{\tealD {\cancel {x}}(x+3)}{\tealD{\cancel x}(x+5)}\\ \\ &=\dfrac{x+3}{x+5} \end{aligned}
Step 4: Final answer
Recall that the original expression is defined for x0,5x\neq 0,-5. The simplified expression must have the same restrictions.
Because of this, we must note that x0x\neq 0. We do not need to note that x5x\neq -5, since this is understood from the expression.
Original expression:
x2+3xx2+5x=x(x+3)x(x+5)\dfrac{x^2+3x}{x^2+5x}=\dfrac{ x(x+3)}{ \blueD x\maroonD{(x+5)}}
Simplified expression:
x+3x+5\dfrac{x+3}{\maroonD{x+5}}
Notice how x=5\maroonC{x=-5} makes both denominators 00, and so it is not a valid input. However, this is implied by each expression.
In contrast, if we didn't indicate x0x\neq0 for the simplified expression, then x=0\blueD{x=0} would be a valid input, since 0+30+5=35\dfrac{\blueD0+3}{\blueD0+5}=\dfrac{3}{5}.
Because the expression evaluated at x=0\blueD{x=0} should be undefined (see original expression), we must restrict it from the domain.
In conclusion, the simplified form is written as follows:
x+3x+5\dfrac{x+3}{x+5} for x0x\neq 0

A note about equivalent expressions

Original expression\quadSimplified expression
x2+3xx2+5x\dfrac{x^2+3x}{x^2+5x}\quadx+3x+5\dfrac{x+3}{x+5} for x0x\neq 0

The two expressions above are equivalent. This means their output values are the same for all possible xx-values. The table below illustrates this for x=2x=2.
Original expression\quadSimplified expression
Evaluation at x=2\purpleC{x=2}(2)2+3(2)(2)2+5(2)=1014=2527=2527=57\begin{aligned}\dfrac{(\purpleC{2})^2+3(\purpleC{2})}{(\purpleC{2})^2+5(\purpleC{2})}&=\dfrac{10}{14}\\\\&=\dfrac{\purpleC{{2}}\cdot 5}{\purpleC{{2}}\cdot 7}\\\\&=\dfrac{\purpleC{\cancel{2}}\cdot 5}{\purpleC{\cancel{2}}\cdot 7}\\\\&=\dfrac{5}{7}\end{aligned}2+32+5=57=57=57=57\begin{aligned}\dfrac{\purpleC{2}+3}{\purpleC{2}+5}&=\dfrac{5}{7}\\\\&\phantom{=\dfrac57}\\\\&\phantom{=\dfrac57}\\\\&\phantom{=\dfrac57}\end{aligned}
NoteThe result is simplified by canceling a common factor of 2\purpleC 2.The result is already simplified because the factor of xx ((in this case x=2)\purpleC{x=2}), was already canceled during the simplification process.

For this reason, the two expressions have the same value for the same input. However, values that make the original expression undefined often break this rule. Notice how this is the case with x=0\purpleC{x=0}.
Original expression\quadSimplified expression (without restriction)
Evaluation at x=0\purpleC{x=0}(0)2+3(0)(0)2+5(0)=00=undefined\begin{aligned}\dfrac{(\purpleC{0})^2+3(\purpleC{0})}{(\purpleC{0})^2+5(\purpleC{0})}&=\dfrac{0}{0}\\\\&=\text{undefined}\end{aligned}0+30+5=35undefined\begin{aligned}\dfrac{\purpleC{0}+3}{\purpleC{0}+5}&=\dfrac{3}{5}\\\\\\\\&\phantom{\text{undefined}}\end{aligned}

Because the two expressions must be equivalent for all possible inputs, we must require x0x\neq 0 for the simplified expression.

Misconception alert

Note that we cannot cancel the xx's in the expression below. This is because these are terms rather than factors of the polynomials!
x+3x+5  \dfrac{x+3}{x+5}~~\Large{\goldD{\neq}}  35~\dfrac{3}{5}
This becomes clear when looking at a numerical example. For example, suppose x=2\purpleC{x=2}.
2+32+5  \dfrac{\purpleC2+3}{\purpleC2+5}~~\Large{\goldD{\neq}} 35~\dfrac{3}{5}
As a rule, we can only cancel if the numerator and denominator are in factored form!

Summary of the simplification process

  • Step 1: Factor the numerator and the denominator.
  • Step 2: List restricted values.
  • Step 3: Cancel common factors.
  • Step 4: Simplify and note any restricted values not implied by the expression.

Check your understanding

1) Simplify 6x+202x+10\dfrac{6x+20}{2x+10}.
Choose 1 answer:
Choose 1 answer:
Step 1: Factor the numerator and denominator
6x+202x+10=2(3x+10)2(x+5)\dfrac{6x+20}{2x+10}=\dfrac{2(3x+10)}{2(x+5)}
Step 2: List restricted values
Here we see that x5\blueD{x\neq{-5}}.
2(3x+10)2(x+5)\dfrac{2(3x+10)}{2\blueD{(x+5)}}
Step 3: Cancel common factors
2(3x+10)2(x+5)=2(3x+10)2(x+5)=3x+10x+5\begin{aligned}\dfrac{\tealD2(3x+10)}{\tealD{{2}}(x+5)}&=\dfrac{\tealD{\cancel{2}}(3x+10)}{\tealD{\cancel{2}}(x+5)}\\ \\ &=\dfrac{3x+10}{x+5} \end{aligned}
Step 4: Final answer
We write the simplified form as follows:
3x+10x+5\dfrac{3x+10}{x+5}
The original expression requires x5x\neq-5. The simplified expression also requires x5x\neq-5. We don't need to restrict any additional values.
2) Simplify x33x24x25x\dfrac{x^3-3x^2}{4x^2-5x}.
for xx\neq
Step 1: Factor the numerator and denominator
x33x24x25x=x2(x3)x(4x5)\dfrac{x^3-3x^2}{4x^2-5x}=\dfrac{x^2(x-3)}{x(4x-5)}
Step 2: List restricted values
Here we see that x0\blueD{x\neq0} and x54\purpleC{x\neq \dfrac54}.
x2(x3)x(4x5)\dfrac{x^2(x-3)}{\blueD x\purpleC{(4x-5)}}
Step 3: Cancel common factors
x2(x3)x(4x5)=xx(x3)x(4x5)=xx(x3)x(4x5)=x(x3)4x5\begin{aligned}\dfrac{x^2(x-3)}{x(4x-5)}&=\dfrac{\tealD x\cdot x(x-3)}{\tealD x(4x-5)}\\ \\ &=\dfrac{\tealD{\cancel{ x}}\cdot x(x-3)}{\tealD{\cancel{ x}}(4x-5)}\\ \\ &=\dfrac{x(x-3)}{4x-5} \end{aligned}
Step 4: Final answer
We write the simplified form as follows:
x(x3)4x5\dfrac{x(x-3)}{4x-5} for x0x\neq 0
The original expression requires x0,54x\neq 0,\dfrac54. We do not need to note that x54x\neq \dfrac54, since this is understood from the expression.

Example 2: Simplifying x29x2+5x+6\dfrac{x^2-9}{x^2+5x+6}

Step 1: Factor the numerator and denominator
x29x2+5x+6=(x3)(x+3)(x+2)(x+3)\dfrac{x^2-9}{x^2+5x+6}=\dfrac{(x-3)(x+3)}{({x+2})({x+3})}
To factor the numerator, we can use the difference of squares pattern:
a2b2=(a+b)(ab)\blueD{ a}^2-\greenD {b}^2=(\blueD {a}+\greenD b)(\blueD {a}-\greenD{ b})
So we have the following:
x29=x232=(x+3)(x3)\begin{aligned}x^2-9&=\blueD{x}^2-\greenD{3}^2\\ \\ &=(\blueD{x}+\greenD3)(\blueD{x}-\greenD3) \end{aligned}
We can factor the denominator using the sum-product pattern.
(x+a)(x+b)= x2+(a+b) x+ab(x+a)(x+b)=\ x^2 +\goldD{(a+b)}\ x + \purpleC{ab}
To factor x2+5x+6x^2+5x+6, we can find factors of 66 that add up to 55. Since 23=62\cdot 3=6 and 2+3=52+3=5, the polynomial factors to (x+2)(x+3)(x+2)(x+3).
Step 2: List restricted values
Since division by 00 is undefined, here we see that x2\blueD{x\neq-2} and x3\purpleC{x\neq -3}.
(x3)(x+3)(x+2)(x+3)\dfrac{(x-3)(x+3)}{\blueD{(x+2)}\purpleC{(x+3)}}
Step 3: Cancel common factors
Notice that the numerator and denominator share a common factor of x+3\tealD{x+3}. This can be canceled out.
(x3)(x+3)(x+2)(x+3)=(x3)(x+3)(x+2)(x+3)=x3x+2\begin{aligned}\dfrac{(x-3)\tealD{(x+3)}}{(x+2)\tealD{(x+3)}}&=\dfrac{(x-3)\tealD{\cancel{(x+3)}}}{(x+2)\tealD{\cancel{(x+3)}}}\\ \\ &=\dfrac{x-3}{x+2} \end{aligned}
Step 4: Final answer
We write the simplified form as follows:
x3x+2\dfrac{x-3}{x+2} for x3x\neq -3
The original expression requires x2,3x\neq-2,-3. We do not need to note that x2x\neq -2, since this is understood from the expression.

Check for understanding

3) Simplify x23x+2x21\dfrac{x^2-3x+2}{x^2-1}.
Choose 1 answer:
Choose 1 answer:
Step 1: Factor the numerator and denominator
x23x+2x21=(x2)(x1)(x+1)(x1)\dfrac{x^2-3x+2}{x^2-1}=\dfrac{(x-2)(x-1)}{(x+1)(x-1)}
Step 2: List restricted values
Here we see that x1\blueD{x\neq-1} and x1\purpleC{x\neq1}.
(x2)(x1)(x+1)(x1)\dfrac{(x-2)(x-1)}{\blueD{(x+1)}\purpleC{(x-1)}}
Step 3: Cancel common factors
(x2)(x1)(x+1)(x1)=(x2)(x1)(x+1)(x1)=x2x+1\begin{aligned}\dfrac{(x-2)\tealD{(x-1)}}{(x+1)\tealD{(x-1)}}&=\dfrac{(x-2)\tealD{\cancel{(x-1)}}}{(x+1)\tealD{\cancel{(x-1)}}}\\ \\ &=\dfrac{x-2}{x+1} \end{aligned}
Step 4: Final answer
We write the simplified form as follows:
x2x+1\dfrac{x-2}{x+1} for x1x\neq 1
The original expression requires x±1x\neq\pm1. We do not need to note that x1x\neq -1, since this is understood from the expression.
4) Simplify x22x15x2+x6\dfrac{x^2-2x-15}{x^2+x-6}.
for xx\neq
Step 1: Factor the numerator and denominator
x22x15x2+x6=(x5)(x+3)(x+3)(x2)\dfrac{x^2-2x-15}{x^2+x-6}=\dfrac{(x-5)(x+3)}{(x+3)(x-2)}
Step 2: List restricted values
Here we see that x3\blueD{x\neq-3} and x2\purpleC{x\neq 2}.
(x5)(x+3)(x+3)(x2)\dfrac{(x-5)(x+3)}{\blueD{(x+3)}\purpleC{(x-2)}}
Step 3: Cancel common factors
(x5)(x+3)(x+3)(x2)=(x5)(x+3)(x+3)(x2)=x5x2\begin{aligned}\dfrac{(x-5)\tealD{(x+3)}}{\tealD{(x+3)}(x-2)}&=\dfrac{(x-5)\tealD{\cancel{(x+3)}}}{\tealD{\cancel{(x+3)}}(x-2)}\\ \\ &=\dfrac{x-5}{x-2} \end{aligned}
Step 4: Final answer
We write the simplified form as follows:
x5x2\dfrac{x-5}{x-2} for x3x\neq -3
The original expression requires x3,2x\neq-3,2. We do not need to note that x2x\neq 2, since this is understood from the expression.

What's next?

You can move on to our advanced article about simplifying rational expressions, where you will see more examples with more challenging cases.
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