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Simplifying rational expressions (advanced)

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 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
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:
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.

Check your understanding

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

Example 2: Simplifying 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.
=(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 x, does not equal, 3 and x, does not equal, minus, 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 minus, 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:
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.

Check your understanding

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.
Choose 1 answer:
Choose 1 answer:

3) Simplify start fraction, 15, minus, 10, x, divided by, 8, x, cubed, minus, 12, x, squared, end fraction.
for x, does not equal
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Let's try some more problems

4) Simplify start fraction, 3, x, divided by, 15, x, squared, minus, 6, x, end fraction.
Choose 1 answer:
Choose 1 answer:

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
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

6) Simplify start fraction, 6, x, squared, minus, 12, x, divided by, 6, x, minus, 3, x, squared, end fraction.
Choose 1 answer:
Choose 1 answer:

Want to join the conversation?

  • piceratops seedling style avatar for user cobri17
    I'm completely lost on Rational expressions. I know how to factor, however I don't understand the whole restriction thing.
    (8 votes)
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    • aqualine tree style avatar for user Judith Gibson
      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.
      (7 votes)
  • blobby green style avatar for user khare9770
    3x+1/x : express as rational expression
    (6 votes)
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  • blobby green style avatar for user Fred Haynes
    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
    (3 votes)
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  • spunky sam blue style avatar for user mohammad ismail memon
    it is pretty hard and specially the one where you have to choose 3 answers
    (4 votes)
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  • aqualine ultimate style avatar for user Tryphosa Thatheu
    (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)
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  • aqualine ultimate style avatar for user 007euclidd
    shouldn't (3-2x) and (2x-3) simplify to (+1) when divided, not (-1) because they are both negative already?
    (1 vote)
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  • blobby green style avatar for user mntambonondumiso24
    How do you simplify (1/x)/1/2rootx
    (1 vote)
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    • cacteye blue style avatar for user Jerry Nilsson
      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.
      (3 votes)
  • leaf blue style avatar for user Igor Skoldin
    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)
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  • blobby green style avatar for user robertalkemade
    in example 4
    3*x/(15*x^2-6*x)
    shouldnt answer = 1/(5*x-2)
    x<>0 AND X<>6/15
    (1 vote)
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  • leaf green style avatar for user Justin Thalib
    Does the lesson called Multiply & divide rational expressions (advanced) require the multiplies answer to be in expanded form of factored form?
    (1 vote)
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