Simplifying rational expressions (advanced)

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  10x32x218x~\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.
10x32x218x=25xx22x(x9)\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 x0{x\neq0} and x9{x\neq9}.
Step 3: Cancel common factors
25xx22x(x9)=25xx22x(x9)=5x2x9\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}
Step 4: Final answer
We write the simplified form as follows:
5x2x9\dfrac{5x^2}{x-9} for x0x\neq 0
Recall that the original expression requires x0,9x\neq 0,9. The simplified expression must have the same undefined values.
Because of this, we must note that x0x\neq 0. We do not need to note that x9x\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.

Check your understanding

1) Simplify 6x212x49x3\dfrac{6x^2}{12x^4-9x^3}.
Choose 1 answer:
Choose 1 answer:
Step 1: Factor the numerator and denominator
6x212x49x3=32x23x3(4x3)\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 x0{x\neq0} and x34{x\neq\dfrac34}.
Step 3: Cancel common factors
32x23x3(4x3)=32x23xx2(4x3)=32x23xx2(4x3)=2x(4x3)\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}
Step 4: Final answer
2x(4x3)\dfrac{2}{x(4x-3)}
Notice that the original requires x0,34x\neq 0,\dfrac34. The simplified expression has the same requirements. We don't need to restrict any additional values.

Example 2: Simplifying  (3x)(x1)(x3)(x+1)~\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, x3x-3 and 3x3-x are related. In fact, we can factor 1-1 out of the numerator to reveal a common factor of x3x-3.
=(3x)(x1)(x3)(x+1)=1(3+x)(x1)(x3)(x+1)=1(x3)(x1)(x3)(x+1)Commutativity\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 x3{x\neq3} and x1{x\neq-1}.
Step 3: Cancel common factors
=1(x3)(x1)(x3)(x+1)=1(x3)(x1)(x3)(x+1)=1(x1)x+1=1xx+1\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 1-1 into the numerator wasn't necessary, but it is common to do so.
Step 4: Final answer
We write the simplified form as follows:
1xx+1\dfrac{1-x}{x+1} for x3x\neq 3
Recall that the original expression requires x3,1x\neq 3,-1. The simplified expression must have the same undefined values.
Because of this, we must note that x3x\neq 3. We do not need to note that x1x\neq -1, since this is understood from the expression.

Main takeaway

The factors x3x-3 and 3x3-x are opposites since 1(x3)=3x-1\cdot (x-3)=3-x.
In this example, we saw that these factors canceled, but that a factor of 1-1 was added. In other words, the factors x3x-3 and 3x3-x canceled to -1\textit{-1}.
In general opposite factors aba-b and bab-a will cancel to 1-1 provided that aba\neq b.

Check your understanding

2) Simplify (x2)(x5)(2x)(x+5)\dfrac{(x-2)(x-5)}{(2-x)(x+5)}.
Choose 1 answer:
Choose 1 answer:
Step 1: Factor the numerator and denominator
The numerator and denominator are already in factored form.
(x2)(x5)(2x)(x+5)\dfrac{(x-2)(x-5)}{(2-x)(x+5)}
Step 2: List restricted values
From the factored form, we see that x2{x\neq2} and x5{x\neq-5}.
Step 3: Cancel opposite factors
We can cancel (x2)\tealD{(x-2)} and (2x)\tealD{(2-x)} to a factor of 1\goldD{-1}.
(x2)(x5)(2x)(x+5)=1(x2)(x5)(2x)(x+5)=1(x5)x+5=5xx+5\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}
Step 4: Final answer
We write the simplified form as follows:
5xx+5\dfrac{5-x}{x+5} for x2x\neq 2
3) Simplify 1510x8x312x2\dfrac{15-10x}{8x^3-12x^2}.
for xx\neq
Step 1: Factor the numerator and denominator
1510x8x312x2=5(32x)4x2(2x3)\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 x0{x\neq0} and x32{x\neq\dfrac32}.
Step 3: Cancel opposite factors
We can cancel (32x)\tealD{(3-2x)} and (2x3)\tealD{(2x-3)} to a factor of 1\goldD{-1}.
5(32x)4x2(2x3)=5(1)(32x)4x2(2x3)=54x2\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}
Step 4: Final answer
We write the simplified form as follows:
54x2\dfrac{-5}{4x^2} for x32x\neq \dfrac32

Let's try some more problems

4) Simplify 3x15x26x\dfrac{3x}{15x^2-6x}.
Choose 1 answer:
Choose 1 answer:
Step 1: Factor the numerator and denominator
3x15x26x=3x3x(5x2)\dfrac{3x}{15x^2-6x}=\dfrac{ 3 x}{ 3x(5x-2)}
Step 2: List restricted values
From the factored form, we see that x0{x\neq0} and x25{x\neq\dfrac25}.
Step 3: Cancel common factors
3x3x(5x2)=3x3x(5x2)=15x2\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 11 still remains.
Step 4: Final answer
We write the simplified form as follows:
15x2\dfrac{ 1}{5x-2} for x0x\neq 0
5) Simplify 3x315x2+12x3x3\dfrac{3x^3-15x^2+12x}{3x-3}.
for xx\neq
Step 1: Factor the numerator and denominator
Notice that there is a common factor of 3x3x in the numerator. We can factor this out, then factor the resulting trinomial.
3x315x2+12x3x3=3x(x25x+4)3(x1)=3x(x4)(x1)3(x1)\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 x1{x\neq1}.
Step 3: Cancel common factors
3x(x4)(x1)3(x1)=3x(x4)(x1)3(x1)=x(x4)\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}
Step 4: Final answer
We write the simplified form as follows:
x(x4)x(x-4) for x1x\neq 1
6) Simplify 6x212x6x3x2\dfrac{6x^2-12x}{6x-3x^2}.
Choose 1 answer:
Choose 1 answer:
Step 1: Factor the numerator and denominator
Notice that there is a common factor of 6x6x in the numerator and 3x3x in the denominator.
6x212x6x3x2=6x(x2)3x(2x)\begin{aligned}\dfrac{6x^2-12x}{6x-3x^2}&=\dfrac{6x(x-2)}{3x(2-x)}\\ \end{aligned}
Step 2: List restricted values
From the factored form, we see that x0x\neq 0 and x2x\neq2.
Step 3: Cancel common factors and opposite factors
Notice that we can write 6x6x as 23x2\cdot 3x. Then we can cancel 3x3x from both the numerator and the denominator.
Next we can cancel the opposite factors (x2)(x-2) and (2x)(2-x) to 1-1.
6x(x2)3x(2x)=2(x2)(2x)=2(1)(x2)(2x)=2\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}
Step 4: Final answer
We write the simplified form as follows:
2-2 for x0,2x\neq 0,2
Loading