Video Transcript
Partial Fractions: Nonrepeated
Linear Factors
In this video, we’re going to
discuss partial fraction decomposition. We’re going to discuss what we mean
by this. And we’re going to focus on the
case where our denominator factors to give us nonrepeated linear factors. We’re going to go through the
entire method step by step. And we’ll discuss what happens when
the degree of the polynomial in our numerator is greater than or equal to the degree
of the polynomial in our denominator. Before we start with partial
fraction decomposition, let’s remember how to add two rational functions
together.
There’s a few different ways of
doing this. For example, we could use cross
multiplication. We multiply the denominator of our
first fraction and the numerator of our second fraction. Then we add onto this the numerator
of our first fraction multiplied by the denominator of our second fraction. This would then give us our new
numerator. Finally, we would then multiply our
two denominators together. This would give us our new
denominator. And this is one way of adding these
two fractions together. However, we’re going to focus on a
different method. To add two fractions together, we
want the denominators of both fractions to be equal. Then we just need to add the
numerators. To do this, we want to find the
lowest common multiple of our denominators. This can often be very tricky. However, we have two unique linear
factors.
So in this case, we just need to
multiply our denominators together. So how does this help us add these
two fractions together? We want the denominator of our
second fraction to be 𝑥 minus one times 𝑥 plus two. So we’ll multiply the entire second
fraction by 𝑥 plus two divided by 𝑥 plus two. Of course, this is the same as
multiplying it by one. But this will allow us to add our
two fractions together. And we also want our first fraction
to have the denominator 𝑥 minus one times 𝑥 plus two. So we’ll multiply our numerator and
our denominator through by 𝑥 minus one. We can now simplify this
expression. Let’s start with our first
fraction. We’ll just multiply the numerator
and the denominator by 𝑥 minus one. This gives us three times 𝑥 minus
one divided by 𝑥 minus one times 𝑥 plus two.
We can do the same with our second
fraction. We multiply the numerator and the
denominator by 𝑥 plus two. This gives us one times 𝑥 plus two
divided by 𝑥 minus one times 𝑥 plus two. And now we can see we’ve made the
denominators of our two fractions equal. So we can just add our numerators
together. Doing this gives us the following
expression. And we can then simplify our
numerator. In the second term in our
numerator, multiplying by one does not change its value.
Next, in the first term in our
numerator, we can distribute three over our parentheses. We see this will give us three 𝑥
minus three. So we’ve now rewritten our fraction
as three 𝑥 minus three plus 𝑥 plus two all divided by 𝑥 minus one times 𝑥 plus
two. And we can then simplify this
further. We have three 𝑥 plus 𝑥 is equal
to four 𝑥 and negative three plus two is equal to negative one. So this gives us four 𝑥 minus one
divided by 𝑥 minus one times 𝑥 plus two.
So let’s think about what we’ve
done. We’ve added our two fractions
together into one new fraction. Partial fraction decomposition will
be a lot like the reverse of this process. We’ll start with one rational
function. We’ll want to factor our
denominator. Then we want to reverse this
process to write it as the sum of two fractions. In particular, we’ll be interested
in making sure that our numerators are constants.
And there’s one thing worth
highlighting about this point. If our numerator is a constant,
then let’s think about what happens to our numerators. We multiply this by the lowest
common multiple of the denominators. We can then ask the question, what
happens to the degree of the polynomial in our numerator compared to the degree of
the polynomial in our denominator?
Well, let’s consider where the
terms in our numerators came from. They’ll be the constants in our
numerators. And remember, we wanted to make the
denominator of each fraction the lowest common multiples of our denominator. So we multiplied each of our
numerators by the remaining part of this lowest common denominator. So our numerator is made up of
polynomials with smaller degree than our denominator. So if we try and reverse this
process in this way, we must have our numerator is a smaller degree polynomial than
our denominator. Otherwise, this won’t work. Let’s now look at some
examples.
The expression two 𝑥 plus one
divided by 𝑥 plus two times 𝑥 plus three can be written in the form 𝐴 divided by
𝑥 plus three plus 𝐵 divided by 𝑥 plus two. Find the values of 𝐴 and 𝐵.
We’re given an expression which is
a linear function divided by a quadratic function, and this quadratic function is
fully factored. We’re told we can write this in the
form 𝐴 divided by 𝑥 plus three plus 𝐵 divided by 𝑥 plus two, where 𝐴 and 𝐵 are
some constants. We need to find the values of 𝐴
and 𝐵. First, the question tells us we can
rewrite this expression in this form, so they must be equal. We just need to find the values of
𝐴 and 𝐵 which make this expression true.
So let’s start by adding these two
expressions together. We can see their denominators are
not equal, so we’ll need to make them equal so we can add these together. To do this, we’ll multiply our
first fraction by 𝑥 plus two divided by 𝑥 plus two. This way, the denominator of our
first fraction will be 𝑥 plus three times 𝑥 plus two, which is worth noting is the
same as the expression we have on the left. And we want to do the same with our
second fraction. We’ll see we’ll need to multiply
this by 𝑥 plus three divided by 𝑥 plus three. This gives us the following
expression.
Now that our denominators are
equal, we can just add the numerators together. This gives us 𝐴 times 𝑥 plus two
plus 𝐵 multiplied by 𝑥 plus three all divided by 𝑥 plus two times 𝑥 plus
three. But now we can see something. The denominator we have on the
left-hand side of our expression and the denominator we have on the right-hand side
of our expression are equal. So for these two functions to be
equal, the numerators must also be equal. In fact, we have something even
stronger. These must be equal for all values
of 𝑥. So equating the two numerators, we
get two 𝑥 plus one is equal to 𝐴 times 𝑥 plus two plus 𝐵 multiplied by 𝑥 plus
three.
It’s worth pointing out here
sometimes you’ll see this written as an equivalence relation. This is to reiterate the fact that
this must be true for all values of 𝑥. However, we’ll just write this as
equals and keep this fact in mind. We now want to solve for our values
of 𝐴 and 𝐵. There are two main methods for
doing this. We could equate the coefficients on
the left- and right-hand side of our equation. But it’s often easier to substitute
values of 𝑥 into this expression. For example, we can see 𝐴 is
multiplied by 𝑥 plus two. So we could eliminate 𝐴 by
multiplying it by zero. In other words, if we substitute 𝑥
is equal to negative two into this expression, we then multiply 𝐴 by zero. We eliminate it from our
equation.
In fact, we can see we can do the
same thing for 𝐵. We substitute 𝑥 is equal to
negative three. Let’s start by substituting 𝑥 is
equal to negative two. We get two times negative two plus
one is equal to 𝐴 multiplied by negative two plus two plus 𝐵 times negative two
plus three. We can then simplify this
expression. First, two times negative two plus
one is equal to negative three. Of course, our coefficient of 𝐴
simplifies to give us zero and our coefficient of 𝐵 simplifies to give us one. So, in fact, this entire equation
simplifies to give us negative three is equal to 𝐵.
We’ll now do the same and eliminate
our variable 𝐵. We’ll substitute 𝑥 is equal to
negative three. This gives us two times negative
three plus one is equal to 𝐴 multiplied by negative three plus two plus 𝐵 times
negative three plus three. If we evaluate this expression,
this time, it gives us negative five is equal to negative 𝐴. We can then multiply both sides of
this equation through by negative one. This gives us that 𝐴 is equal to
five. Therefore, we were able to show for
the expression given to us in the question to be written in the form 𝐴 over 𝑥 plus
three plus 𝐵 over 𝑥 plus two, we must have 𝐴 is equal to five and 𝐵 is equal to
negative three.
In this question, we were able to
take our expression and write it in a new form by using partial fraction
decomposition. However, we were given the form we
need to write it in. We now need to find out how we
would approach this problem if we’re not given the form to write it in.
Express one over 𝑥 squared minus
one in partial fractions.
The question wants us to express
one over 𝑥 squared minus one in partial fractions. There are two things we always need
to do when we’re asked to use partial fractions. The first thing we need to do is
check the degrees of the polynomial in our numerator and our denominator, since we
can only use partial fractions if the degree of the polynomial in our numerator is
smaller than the one in the denominator. In this case, the numerator is a
constant. So its highest power of 𝑥 is 𝑥 to
the zeroth power. So we say its degree is zero. And in our denominator, the term
with the highest power of 𝑥 is 𝑥 squared. So we say its degree is two. So we’ve done our first check.
The second thing we need to do is
fully factor our denominator. In our case, our denominator is a
difference between squares, so we can factor this as 𝑥 minus one times 𝑥 plus
one. We’re now ready to start using
partial fraction decomposition. First, we’ll write our expression
as one divided by 𝑥 minus one times 𝑥 plus one. In this case, we can see we have
two unique factors in our denominator. So we’ll try and write this in the
form 𝐴 divided by 𝑥 minus one plus 𝐵 divided by 𝑥 plus one, where 𝐴 and 𝐵 are
constants. We want to try and find our values
of 𝐴 and 𝐵, so we’re going to need to add these two fractions together.
To add these two factions together,
their denominators need to be equal. So we’ll multiply our first
fraction by 𝑥 plus one divided by 𝑥 plus one and our second fraction by 𝑥 minus
one divided by 𝑥 minus one. This gives us the following
expression. Now that their denominators are
equal, to add these two terms together, we just need to add the numerators. Adding the numerators of these two
terms together, we get 𝐴 times 𝑥 plus one plus 𝐵 multiplied by 𝑥 minus one
divided by 𝑥 minus one times 𝑥 plus one. But now we can see the denominator
of the expression on the left-hand side of our equation and the denominator of the
expression on the right-hand side of our equation are equal.
So the only way for these two
functions to be equal is if their numerators are equal. So let’s equate the numerators on
both sides of the equation. This gives us one is equal to 𝐴
times 𝑥 plus one plus 𝐵 times 𝑥 minus one. We could find our values of 𝐴 and
𝐵 by equating coefficients. However, remember, this must be
true for any value of 𝑥. In particular, we could substitute
𝑥 is equal to negative one. Then, our coefficient of 𝐴 becomes
zero. So we eliminate it from the
equation. Substituting 𝑥 is equal to
negative one, we get one is equal to 𝐴 times negative one plus one plus 𝐵
multiplied by negative one minus one.
Of course, as we said earlier, our
coefficient of 𝐴 simplifies to give us zero. So this simplifies to give us that
one is equal to negative two 𝐵. And we can divide this equation
through by negative two to get that 𝐵 is equal to negative one-half. We can do something similar to
eliminate our variable 𝐵. We substitute 𝑥 is equal to
one. This will make the coefficient of
𝐵 equal to zero. Substituting 𝑥 is equal to one, we
get one is equal to 𝐴 times one plus one plus 𝐵 multiplied by one minus one. This time, our coefficient of 𝐵
simplifies to give us zero. So we can simplify this equation to
get one is equal to two 𝐴. And we divide this equation through
by two to see that 𝐴 is equal to one-half.
The last thing we’ll do is
substitute our value of 𝐴 is equal to one-half and our value of 𝐵 is equal to
negative one-half into our expression in partial fractions. So we want to substitute 𝐴 is
equal to one-half and 𝐵 is equal to negative one-half into this expression. Let’s start with 𝐴 is equal to
one-half. However, we could simplify this
expression. Instead of having one-half in our
numerator, we’ll have one in our numerator and two in our denominator.
We now want 𝐵 is equal to negative
one-half. And we can do something
similar. Instead of having one-half in our
numerator, we’ll have one in our numerator and two in our denominator. However, since our numerator is
negative, we’re going to need to subtract this value, and this is our final
answer. Therefore, by using partial
fractions, we were able to express one over 𝑥 squared minus one as one over two
times 𝑥 minus one minus one over two times 𝑥 plus one.
So far, our examples have had two
unique factors in the denominator. Let’s see what would happen if
instead we had three unique factors in the denominator.
Express 𝑥 squared minus two all
over 𝑥 plus two times 𝑥 minus three multiplied by 𝑥 plus one in partial
fractions.
We need to write this expression in
partial fractions. And the first thing we always do is
check the degree of the polynomial in our numerator and check the degree of the
polynomial in our denominator. In our numerator, the highest power
of 𝑥 is the term 𝑥 squared. So the degree of the polynomial in
our numerator is two. In our denominator, when we
multiply out our parentheses, we’ll see we’ll have 𝑥 cubed.
So the degree of the polynomial in
our denominator is three. So our denominator is a
higher-degree polynomial than our numerator. This means we can proceed with
partial fractions. We see we have three unique roots
in our denominator. And when our denominator has three
unique roots, we need to split our fraction into three terms. By partial fractions, we get that
our fraction is equal to 𝐴 over 𝑥 plus one plus 𝐵 over 𝑥 plus two plus 𝐶 over
𝑥 minus three for some constants 𝐴, 𝐵, and 𝐶.
We now want to add these three
terms together. To do this, we need their
denominators to match. Let’s start with our first
term. Our first term has the denominator
𝑥 plus one. So we’ll multiply the numerator and
the denominator of this fraction by 𝑥 plus two times 𝑥 minus three. So we multiply 𝐴 over 𝑥 plus one
by 𝑥 plus two times 𝑥 minus three divided by 𝑥 plus two times 𝑥 minus three. We now want to rewrite the
denominator in our second term in the same way. We’ll multiply the numerator and
the denominator of this fraction by 𝑥 plus one times 𝑥 minus three. This time, for our second term, we
get 𝐵 over 𝑥 plus two times 𝑥 plus one times 𝑥 minus three divided by 𝑥 plus
one multiplied by 𝑥 minus three. We want to do the same with our
third and final term. This time, we need to multiply the
numerator and the denominator by 𝑥 plus one times 𝑥 plus two. And this gives us the following
expression.
Now that we’ve rewritten our
denominators to be equal, we can just add these terms together by adding the
numerators together. When we do this, we get 𝐴 times 𝑥
plus two times 𝑥 minus three plus 𝐵 multiplied by 𝑥 plus one multiplied by 𝑥
minus three plus 𝐶 times 𝑥 plus one times 𝑥 plus two all divided by 𝑥 plus one
times 𝑥 plus two times 𝑥 minus three. Now we can see the denominator on
the left-hand side of our equation and the denominator on the right-hand side of our
equation are equal. So for these two functions to be
equal, the numerators must also be equal.
This means we can just equate our
numerators. We get 𝑥 squared minus two is
equal to 𝐴 times 𝑥 plus two times 𝑥 minus three plus 𝐵 multiplied by 𝑥 plus one
times 𝑥 minus three plus 𝐶 times 𝑥 plus one multiplied by 𝑥 plus two. And remember, this has to be true
for all values of 𝑥. So we can eliminate variables by
substituting different values of 𝑥. For example, if we substitute 𝑥 is
equal to negative two, then our coefficient of 𝐴 and our coefficient of 𝐶 are
equal to zero.
Substituting 𝑥 is equal to
negative two, on the left-hand side of our equation, we get negative two squared
minus two. And remember, our coefficients of
𝐴 and 𝐶 will be zero, so we only need to substitute 𝑥 is equal to negative two to
find our coefficient of 𝐵. So we’ll substitute 𝑥 is equal to
negative two to find our coefficient of 𝐵. We get 𝐵 times negative two plus
one multiplied by negative two minus three. Simplifying this equation, we get
two is equal to five 𝐵. And then we just divide through by
five to see that 𝐵 is equal to two-fifths.
We can do something similar to find
the value of 𝐶. This time, we’ll substitute 𝑥 is
equal to three. Now when we substitute this in, our
coefficients of 𝐴 and 𝐵 will both be equal to zero. Substituting 𝑥 is equal to three
and using the fact that our coefficients of 𝐴 and 𝐵 will be equal to zero, we get
three squared minus two is equal to 𝐶 times three plus one times three plus
two. Simplifying this expression, we get
that seven is equal to 20𝐶, and then we can just solve this. We see that 𝐶 is equal to seven
divided by 20.
Finally, we do exactly the same to
find our value of 𝐴. This time we’re going to substitute
𝑥 is equal to negative one. Doing this and remembering our
coefficients of 𝐵 and 𝐶 will now be equal to zero, we get negative one squared
minus two is equal to 𝐴 times negative one plus two multiplied by negative one
minus three. And then we simplify and solve this
equation and we get that 𝐴 is equal to one-quarter.
The last thing we need to do to
solve this equation is to substitute our values of 𝐴, 𝐵, and 𝐶 into our
expression in partial fractions. So, by using 𝐴 is one-quarter, 𝐵
is two-fifths, and 𝐶 is seven over 20, we were able to express 𝑥 squared minus two
divided by 𝑥 plus two times 𝑥 minus three times 𝑥 plus one in partial
fractions. We got the expression one over four
times 𝑥 plus one plus two over five times 𝑥 plus two plus seven over 20 times 𝑥
minus three.
The last thing we need to talk
about is what happens if the polynomial in our numerator is of an equal or higher
degree than the polynomial in our denominator.
Express 𝑥 squared minus five
divided by 𝑥 squared minus four in a form which we can rewrite into partial
fractions.
We’re given a quadratic divided by
a quadratic. Since our numerator has an equal
degree to our denominator, we can’t directly write this in partial fractions. So we’ll have to do something
different. We’ll rewrite this by using
algebraic division. We see that 𝑥 squared minus four
goes into 𝑥 squared minus five one time. So it goes in once and then we need
to subtract one times 𝑥 squared minus four. We have 𝑥 squared minus 𝑥 squared
is equal to zero, and then negative five minus negative four is equal to negative
one.
So 𝑥 squared minus four goes into
𝑥 squared minus five once with a remainder of one. But what does this actually
mean? When we divide 𝑥 squared minus
five by 𝑥 squared minus four, we get one and a remainder of one. So it’s equal to one plus one over
𝑥 squared minus four. And then, this is one plus a
constant divided by a quadratic. So we can rewrite this by using
partial fractions.
Let’s go over the key points of
this video. The first thing we do when we’re
using partial function decomposition is to check the degrees of the polynomial in
our numerator and denominator. And if our numerator is of a higher
or equal degree to our denominator, we can rewrite this fraction by using algebraic
division. The next thing we need to do is
factor the denominator fully. Then if all of our factors are
unique and linear, we just write this as a sum of constants divided by each of our
factors. Finally, we just add these
fractions together and solve for our constants by substitution or by equating
coefficients.
