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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 $~\dfrac{10x^3}{2x^2-18x}$space, start fraction, 10, x, cubed, divided by, 2, x, squared, minus, 18, x, end fraction

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.
start fraction, 10, x, cubed, divided by, 2, x, squared, minus, 18, x, end fraction, equals, start fraction, 2, dot, 5, dot, x, dot, x, squared, divided by, 2, dot, x, dot, left parenthesis, x, minus, 9, right parenthesis, end fraction
Step 2: List restricted values
From the factored form, we see that x, does not equal, 0 and x, does not equal, 9.
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:
start fraction, 5, x, squared, divided by, x, minus, 9, end fraction for x, does not equal, 0

### Main takeaway

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

1) Simplify start fraction, 6, x, squared, divided by, 12, x, start superscript, 4, end superscript, minus, 9, x, cubed, end fraction.

## Example 2: Simplifying $~\dfrac{(3-x)(x-1)}{(x-3)(x+1)}$space, start fraction, left parenthesis, 3, minus, x, right parenthesis, left parenthesis, x, minus, 1, right parenthesis, divided by, left parenthesis, x, minus, 3, right parenthesis, left parenthesis, x, plus, 1, right parenthesis, end fraction

Step 1: Factor the numerator and denominator
While it does not appear that there are any common factors, x, minus, 3 and 3, minus, x are related. In fact, we can factor minus, 1 out of the numerator to reveal a common factor of x, minus, 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, does not equal, 3 and x, does not equal, minus, 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 minus, 1 into the numerator wasn't necessary, but it is common to do so.
We write the simplified form as follows:
start fraction, 1, minus, x, divided by, x, plus, 1, end fraction for x, does not equal, 3

### Main takeaway

The factors x, minus, 3 and 3, minus, x are opposites since minus, 1, dot, left parenthesis, x, minus, 3, right parenthesis, equals, 3, minus, x.
In this example, we saw that these factors canceled, but that a factor of minus, 1 was added. In other words, the factors x, minus, 3 and 3, minus, x canceled to start text, negative, 1, end text.
In general opposite factors a, minus, b and b, minus, a will cancel to minus, 1 provided that a, does not equal, b.

2) Simplify start fraction, left parenthesis, x, minus, 2, right parenthesis, left parenthesis, x, minus, 5, right parenthesis, divided by, left parenthesis, 2, minus, x, right parenthesis, left parenthesis, x, plus, 5, right parenthesis, end fraction.

3) Simplify start fraction, 15, minus, 10, x, divided by, 8, x, cubed, minus, 12, x, squared, end fraction.
for x, does not equal

## Let's try some more problems

4) Simplify start fraction, 3, x, divided by, 15, x, squared, minus, 6, x, end fraction.

5) Simplify start fraction, 3, x, cubed, minus, 15, x, squared, plus, 12, x, divided by, 3, x, minus, 3, end fraction.
for x, does not equal

6) Simplify start fraction, 6, x, squared, minus, 12, x, divided by, 6, x, minus, 3, x, squared, end fraction.

## Want to join the conversation?

• I'm completely lost on Rational expressions. I know how to factor, however I don't understand the whole restriction thing.
• If we divide out a common factor from both numerator and denominator so that it disappears entirely from the fraction, then we need to restrict x from being whatever value made that factor equal zero.
For example, ( x + 3 ) ( x - 5 ) / (x + 3) = ( x + 5 ) but x can't equal -3.
For example, ( y - 7 ) ( y + 2 ) / ( y - 7 ) ^2 = ( y + 2 ) / ( y - 7 ), but there's no need to list any restrictions for y because even though one factor of ( y - 7 ) has been divided out, a factor of ( y - 7 ) still remains in the denominator.
• 3x+1/x : express as rational expression
• When we multiply the 3x term by x/x to get a common denominator, we get: (3x^2 + 1)/x
• I just finished an the Practice Problems in the next exercise. To get to the part I don't understand I will simply ask this question which was the last section of the problem.

The final answer was -(z+11)/(z-4) then it was multiplied out to z+11/4-z. Why was the answer -(z+11)/(z-4) not correct.

Thanks
• Your answer is correct. For some reason, KA is setup so it doesn't except some small variations in correct answers. You may want to report it as a problem within the exercise so maybe they'll change it.
• it is pretty hard and specially the one where you have to choose 3 answers
• I totally get it
(1 vote)
• For question 3, I don't understand why the answer is -5/4x^2 for x =/= 3/2. Why not x=/= 0 as well? I got the result (-5/4x^2) correct but when I tried to enter for x=/= "0, 3/2" I keep getting, "we couldn't understand your answer."
• You're right. Since 4x^2 is in the denominator of the answer, I didn't think to include x != 0. And neither did KA, since it's still in the answer, so "everybody knows".
• (x−3)(x+6) x 2 −16 ​ ÷ x 2 +cx−18 (x+a)(a−x) ​ =−1

I want to know what c is..
(1 vote)
• Hint: Multiply out: (x-3)(x+6) to find the value of "c"
• Determine whether 3^{2m}\cdot 4^{2m}3
2m
⋅4
2m
3, start superscript, 2, m, end superscript, dot, 4, start superscript, 2, m, end superscript is equivalent to each of the following expressions.
• shouldn't (3-2x) and (2x-3) simplify to (+1) when divided, not (-1) because they are both negative already?
(1 vote)
• 3-2x and 2x-3 are not both negative. If one is negative, then the other is positive, since 2x-3 is -1·(3-2x). At best, they may both be 0 (if x=3/2), but then we can't divide them, since we can't divide by 0.
• How do you simplify (1/x)/1/2rootx
(1 vote)
• If this expression is correctly written, then the order of operations (PEMDAS) tells us to first simplify 1/𝑥, which cannot be simplified any further, then √𝑥 (because this is technically an exponent of 𝑥), which can neither be simplified any further, which means that we keep these expressions as they are.
Next is multiplication/division, which we perform from left to right:
(1/𝑥)/1 = 1/𝑥.
(1/𝑥)/2 = 1/(2𝑥).
(1/(2𝑥)) ∙ √𝑥 = √𝑥/(2𝑥) = 1/(2√𝑥)
And we're done.

– – –

However, I have a hunch that you meant to simplify (1/𝑥)/(1/(2√𝑥)), where we can't simplify any of the parentheses or exponents, but we do have a fraction of the type (1/𝑎)/(1/𝑏), which simplifies as 𝑏/𝑎.
So, (1/𝑥)/(1/(2√𝑥)) = 2√𝑥/𝑥 = 2/√𝑥.

– – –

Note that, in both examples, the domain of the simplified expression is the same as the domain of the original expression, which is not always the case. If the domain is somehow changed during the simplification process, we should include the domain of the original expression in our answer.
• What would happen if in the Example 2 we multiply a term in the denominator (instead of a term in the nominator) to get a common factor?

(3-x)(x-1)/(x-3)(x+1)
= (3-x)(x-1)/(-1)(3-x)(x+1) # multiply (x-3) by -1 and get the common factor of (3-x)
= (x-1)/(1+x) # cancel out (3-x) and multiply (x+1) by -1 in the denominator

So we get (x-1)/(1+x) for x ≠ 3 (and x ≠ -1, which is implied by the expression). But we should have gotten (1-x)/(1+x), which is not exactly the same. Actually, it is equivalent to (x-1)(-1)/(1+x).

Did I make a mistake or why doesn't it work?
(1 vote)
• You make a mistake here:
"multiply (x+1) by -1"
You say the result is (1+x), but that's not correct. (-1)·(x+1) = (-x - 1)