Step-by-step explanation:
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 \dfrac{x+2}{x+1}
x+1
x+2
start fraction, x, plus, 2, divided by, x, plus, 1, end fraction is all real numbers except \textit{-1}-1start text, negative, 1, end text, or x\neq -1x
=−1x, does not equal, minus, 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 \dfrac 68
8
6
start fraction, 6, divided by, 8, end fraction is \dfrac{3}{4}
4
3
start fraction, 3, divided by, 4, end fraction. Notice how we canceled a common factor of 222 from the numerator and the denominator:
\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}
8
6
=
2⋅4
2⋅3
=
2
⋅4
2
⋅3
=
4
3
Factor
Cancel common factors
Simplify
Example 1: Simplifying \dfrac{x^2+3x}{x^2+5x}
x
2
+5x
x
2
+3x
start fraction, x, squared, plus, 3, x, divided by, x, squared, plus, 5, x, end fraction
Step 1: Factor the numerator and denominator
The only way to see if the numerator and denominator share common factors is to factor them!
\dfrac{x^2+3x}{x^2+5x}=\dfrac{ x(x+3)}{ x(x+5)}
x
2
+5x
x
2
+3x
=
x(x+5)
x(x+3)
start fraction, x, squared, plus, 3, x, divided by, x, squared, plus, 5, x, end fraction, equals, start fraction, x, left parenthesis, x, plus, 3, right parenthesis, divided by, x, left parenthesis, x, plus, 5, right parenthesis, end fraction
Step 2: List restricted values
At this point, it is helpful to notice any restrictions on xxx. These will carry over to the simplified expression.
Since division by 000 is undefined, here we see that \blueD{x\neq0}x
=0start color #11accd, x, does not equal, 0, end color #11accd and \purpleC{x\neq -5}x
=−5start color #aa87ff, x, does not equal, minus, 5, end color #aa87ff.
\dfrac{ x(x+3)}{ \blueD x\purpleC{(x+5)}}
x(x+5)
x(x+3)
start fraction, x, left parenthesis, x, plus, 3, right parenthesis, divided by, start color #11accd, x, end color #11accd, start color #aa87ff, left parenthesis, x, plus, 5, right parenthesis, end color #aa87ff, end fraction
Step 3: Cancel common factors
Now notice that the numerator and denominator share a common factor of xxx. This can be canceled out.
\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}
x(x+5)
x(x+3)
=
x
(x+5)
x
(x+3)
=
x+5
x+3
Step 4: Final answer
Recall that the original expression is defined for x\neq 0,-5x
=0,−5x, does not equal, 0, comma, minus, 5. The simplified expression must have the same restrictions.
Because of this, we must note that x\neq 0x
=0x, does not equal, 0. We do not need to note that x\neq -5x
=−5x, does not equal, minus, 5, since this is understood from the expression. [Can you elaborate here?]
In conclusion, the simplified form is written as follows:
\dfrac{x+3}{x+5}
x+5
x+3
start fraction, x, plus, 3, divided by, x, plus, 5, end fraction for x\neq 0x
=0x, does not equal, 0
A note about equivalent expressions
Original expression\quadSimplified expression
\dfrac{x^2+3x}{x^2+5x}
x
2
+5x
x
2
+3x
start fraction, x, squared, plus, 3, x, divided by, x, squared, plus, 5, x, end fraction\quad\dfrac{x+3}{x+5}
x+5
x+3
start fraction, x, plus, 3, divided by, x, plus, 5, end fraction for x\neq 0x
=0x, does not equal, 0
The two expressions above are equivalent. This means their output values are the same for all possible xxx-values. The table below illustrates this for x=2x=2x, equals, 2.
Original expression\quadSimplified expression
Evaluation at \purpleC{x=2}x=2start color #aa87ff, x, equals, 2, end color #aa87ff\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)
2
+5(2)
(2)
2
+3(2)
=
14
10
=
2⋅7
2⋅5
=
2
⋅7
2
⋅5
=
7
5
\begin{aligned}\dfrac{\purpleC{2}+3}{\purpleC{2}+5}&=\dfrac{5}{7}&\phantom{=\dfrac57}&\phantom{=\dfrac57}&\phantom{=\dfrac57}\end{aligned}
2+5
2+3
=
7
5
=
7
5
=
7
5
=
7
5
