Rational Expressions
Remember rational numbers? The rational numbers are all real numbers that can be written as a fraction. Similarly, rational expressions are
expressions that can be written as a fraction. For example,
2
,
3
1
,
x+1
and
x2 + 2x + 1
2x + 2
are rational expressions because they are written as a fraction. Typically, we
talk about rational expressions where the numerator and denominator are
both polynomials.
The Domain of Rational Expressions
Have you ever typed “1 / 0” into your calculator? If you have, it says: “ERROR: ILLEGAL DIVISION BY ZERO.” This is because zero can NOT be
in the denominator of ANY expression. The same thing goes for rational expressions that are quotients of polynomials. Therefore, we have a restriction
on the values we can plug in to any rational expression. Specifically, we can
not have a denominator of zero. This means that the expression is undefined
for whatever x values make the denominator equal to zero. In general, the
domain of an algebraic expression is the set of all real numbers for which
the expression is defined. Since rational expressions are undefined when the
denominator is zero, the domain of a rational expression is the set of all real
numbers such that the denominator is not zero.
For example, what is the domain of
x2 − 4
x2 + 4x + 2
From the definition, it is the set of all real so that x2 + 4x + 2 6= 0. Since this
is hardly ever zero, it will be easier to find when it is equal to zero. Then
the domain is all real numbers except those that make the denominator zero.
Thankfully, we can factor the denominator into
x2 + 5x + 6 = (x + 2)(x + 3) = 0
when
x = −3 or x = −2
Therefore, our domain is the set of all real numbers x such that x does
not equal -2 and x does not equal -3. We can write this is set notation as:
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{x ∈ R|x 6= −3 and x 6= −2}. This is also sometimes written as {x|x is a real
number and x 6= −3 and x 6= −2}. Sometimes, “:” is used as the separator
instead of “|”, it just depends on the book.
Manipulating Rational Expressions
Rational expressions show up in many places in algebra, and it is very important to be comfortable working with them. This means:
• simplify rational expressions to make them easier to work with
• add and subtract rational expressions to look for more simplifications
• multiply and divide rational expressions to combine them for further
simplification
All of these skills are necessary to effectively work with rational exressions.
These expressions can become cumbersome very quickly, and it is essential
to have these skills mastered in order to use them as tools to solve other
problems rather than the first step you didn’t see.
Simplifying Rational Expressions
Simplifying rational expressions is all about multiplying by 1. The trick is
to find all of these factors of 1 hidden in the problem. First, recall that for
any real number x 6= 0,
x
=1
x
Therefor, whenever we have a factor in the numerator and denominator that
are the same, we can “cancel” the terms. Consider the example used in the
previous section:
x2 − 4
x2 − 4
(x + 2)(x − 2)
x+2 x−2
x−2
x−2
=
=
=
·
= 1·
=
2
x + 4x + 2
(x + 2)(x + 3)
(x + 2)(x + 3)
x+2 x+3
x+3
x+3
Therefore, we have simplified the domain example:
x2 − 4
x−2
=
x2 + 4x + 2
x+3
2
However, the domain of the simplified expression is {x ∈ R|x 6= −3}, which
is NOT the same as the original domain. This is because the two expressions
are not quite equal. They are equal when BOTH are defined. In other words,
they are equal for all x that are in BOTH domains. These expressions are
called equivalent expressions because they are not truly equal, but they
have the same value for all numbers that are in both domains. These two
x−2
domains are different because the domain of x+3
has -2 in its domain, but the
x2 −4
original expression x2 +4x+2 is not defined for this value because the denominator is zero. This leads to an important point. WHEN YOU SIMPLIFY
AN EXPRESSION, YOU DO NOT CHANGE THE DOMAIN!
Arithmetic Operations Involving Rational Expressions
Rational expressions work just like fractions. You add, subtract, multiply
and divide rational expressions the same way you do fractions.
Multiplication is the easiest operation to carry out on fractions. Simply multipily the numerators to get the new numerator, and multiply the
denominators to get the new denominator. For example:
a c
a·c
· =
b d
b·d
In other words, just multiply straight across to get the answer.
Division is almost as simple as multiplication. In fact, you can turn a
division problem with fractions into a multiplication problem. For a longwinded example, consider:
a c
÷ =
b d
=
a·d
b·c
c·d
c·d
a
b
c
d
=
=
a·d
b·c
1
a
b
c
d
·1=
=
a
b
c
d
·
d
c
d
c
=
a·d
b·c
c·d
d·c
a·d
a d
= ·
b·c
b c
a d
a c
÷ = ·
b d
b c
Ok. That was long and laborious. However, we did not break any of the rules
(provided c 6= 0 and d 6= 0, which would make no sense from the beginning
since the initial problem would have division by zero). The bottom line of
that was to show that we can change a division problem into a multiplication
Therefore
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problem if we use the reciprocal of the term we are dividing by. In other
words, we just “flip the second and multiply.” For a concrete example with
numbers:
1 5
1·5
5
1 2
÷ = · =
=
3 5
3 2
3·2
6
Of course, we would have to reduce this answer if we could, such as in this
example:
2 4
2 7
2·7
14
2·7
2 7
7
÷ = · =
=
=
= · =
3 7
3 4
3·4
12
2·6
2 6
6
We had to reduce this final answer. It is always good to reduce the final
answer if it is possible.
When you want to add or subtract fractions, you need a common denominator. Of course, it makes life easier if you have the least common
denominator of the expression. However, it is not completely necessary to
find the least commond denominator because you can simplify the expression after you perform the operation. It just may be easier to find the least
common denominator at the beginning so you do not have to worry about
simplifying a complicated expression at the end. For example, say we want
to subtract
1 1
−
4 2
We can see that 4 = 2 · 2, so the least common denominator is 4. Then, we
can solve this:
1 1
1 1
1 1 2
1 2
1
− = − ·1= − · = − =−
4 2
4 2
4 2 2
4 4
4
However, we can add general fractions by forcing the denominators to be
common denominators (which may not be the least common denominator).
This is accomplished by multiplying both terms by 1, except we use a specific
way of writing it. We multiply by 1 as xx where x is the denominator of the
other term:
a c
a
c
a d c b
a·d c·b
a·d c·b
a·d+c·b
+ = ·1+ ·1 = · + · =
+
=
+
=
b d
b
d
b d d b
b·d d·b
b·d b·d
b·d
Sometimes this is called “cross multiplying” because we end up with a numerator of the sum of the terms you would get if you made an X between the
fractions and multiplied the terms opposite each other, and the denominator
is the product of the denominators:
a c
ad + bc
X =
b d
bd
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What we have here is ad from the \ part of the X, and bc from the / part of
the X. Then, the denominator is the product of the old denominators.
We can easily write any subtraction as an addition problem as:
a
c
a −1 c
a −1 · c
a −c
a c
− = + (−1) = +
· = +
= +
b d
b
d
b
1 d
b
1·d
b
d
So, we now know how to add, subtract, multiply and divide fractions.
Thankfully, rational expressions are nothing other than fractions. The
bad thing is that we do not know the numbers that are the numerator and
denominator because they depend on some variable. However, all we have to
worry about is not dividing by zero. For this reason, we have to make sure
that we take note of the x values that will make the denominator of any of
the terms in the original expression zero. Then, we say the expression is not
defined for these x values by specifying our domain, as stated earlier. If you
know how to add, subtract, multiply and divide fractions, you know how to
add, subtract, multiply and divide rational expressions.
However, the problem comes in when we are faced with the task of reducing the fraction. It is a good idea to factor the original expressions
completely before trying to perform any operations. When multiplying (and
dividing, since it is essentially a multiplication), it is a good idea not to multiply the terms out. After an addition or subtraction, it is a good idea to try
to factor again. Doing these makes it easier to:
• find the domain of the expression
• find the least common denominator of an expression
• simplify the expression by cancelling terms
Now that we have an idea of how to attack these problems, let’s do some
examples.
x2 − 4
x2 − 4x + 3
÷
x2 + 2x + 4
x4 − 9x2
First, we factor all of the expressions:
(x − 2)(x + 2)
(x − 3)(x − 1)
÷ 2
2
(x + 2)
x (x − 3)(x + 3)
Now, we need to figure out the domain. From the first term, we know we
can not have x + 2 = 0 i.e. x 6= −2. In the second term, we can not have
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a denominator of zero, so x 6= 0, x 6= 3 and x 6= −3. Finally, since we are
dividing by the entire second term, it can not equal zero. Therefore, the
numerator of the second term can not be zero either. This happens when
x = 3 or x = 1. Putting it all together, our domain is {x ∈ R such that
x 6= −3 and x 6= −2 and x 6= 0 and x 6= 1 and x 6= 3}. Next, we simplify:
(x − 1) x − 3
(x − 2)(x + 2)
(x − 3)(x − 1)
x−2x+2
÷ 2
=
÷
=
(x + 2)2
x (x − 3)(x + 3)
x+2x+2
x2 (x + 3) x − 3
(x − 1)
x−2
(x − 1)
x−2
·1÷
·1 =
÷ 2
2
x+2
x (x + 3)
x + 2 x (x + 3)
We still have not done anything except find the domain and simplify. Now,
we can “flip the second and multiply.”
x − 2 (x − 1)
x − 2 x2 (x + 3)
(x − 2)x2 (x + 3)
x2 (x − 2)(x + 3)
÷ 2
=
·
=
=
x + 2 x (x + 3)
x+2 x−1
(x + 2)(x − 1)
(x + 2)(x − 1)
At this point, we try to simplify. However, there are no common factors in
the numerator and denominator, so we can not do anything else. We can
either leave this as it is, or multiply it out. We will leave this as it is.
Now, consider a more complicated example.
1
+ y12
x
x
− xy
y
This problem incorporates all of the operations of fractions. However, there
is not much we can do to this. Order of operations tells us to do parenthases
then exponents. Niether of those appear in this equation. Then, we are supposed to multiply and divide next. However, we do not know the reciprocal
of the entire denominator. Therefore, we have to simplify it. We can work
with the numerator and the denominator and try to make each of them one
fraction by performing the addition or subtraction:
1
+ y12
x
x
− xy
y
=
1 y2
·
x y2
x x
·
y x
+
−
1
·x
y2 x
y y
·
x y
=
y2
xy 2
x2
xy
+
−
x
xy 2
y2
xy
=
x+y 2
xy 2
2
x −y 2
xy
Since we have not cancelled any terms, we have not changed the domain
yet. So, we will get the domain from this expression. We know that xy 2
can not equal zero, and xy can not equal zero. Finally, since the entire big
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denominator can not be zero, x2 − y 2 can not equal zero. We know from the
first two conditions that x 6= 0 and y 6= 0. Now, lets find when x2 − y 2 = 0.
This happens when x2 = y 2 , or when x = ±y (because (−a)2 = a2 for all
real numbers a). Therefore, our domain is {x ∈ R such that x 6= 0, y 6= 0
and x 6= ±y}.
The expression can now be rewritten as
x+y 2
xy 2
2
x −y 2
xy
=
x + y2
xy
x + y 2 x2 − y 2
÷
=
·
xy 2
xy
xy 2
x2 − y 2
Now, we can factor and then multiply:
x + y2
xy
x + y2
xy
(xy)(x + y 2 )
·
=
·
=
xy 2
x2 − y 2
xy 2
(x + y)(x − y)
xy 2 (x + y)(x − y)
Now, we can simplify this by canceling xy:
(xy)(x + y 2 )
(x + y 2 )
xy
=
·
=
2
xy (x + y)(x − y)
y(x + y)(x − y) xy
(x + y 2 )
(x + y 2 )
·1=
y(x + y)(x − y)
y(x + y)(x − y)
At this point, there is no further simplification possible. We can not factor
the numerator any more, so we will not be able to cancel with any of the
terms in the denominator.
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