### Video Transcript

In this video, weβll learn how to
decompose rational expressions into partial fractions when the denominator has
repeated linear factors. We begin by recalling the fact that
a fraction or a rational function with two or more distinct linear factors in its
denominator and whose denominator has a higher order than the polynomial in its
numerator can be split into two or more separate fractions with linear
denominators. We essentially reverse the process
of adding algebraic fractions to achieve this, either by using substitution or
equating coefficients.

We use partial fraction
decomposition in integration and in evaluating binomial expansions. So itβs a really important skill to
master. Further, the method of partial
fractions can be used when there are repeated linear factors in the denominator,
such as in the case of two π₯ plus one over π₯ minus three times π₯ plus one cubed,
though it does look slightly different. When dealing with a repeated linear
factor, here thatβs π₯ plus one since itβs being cubed, we follow the usual steps
for dealing with nonrepeated linear factors. This time, though, if π is the
exponent of the repeated linear factor on the denominator, we write π rational
expressions, each of which has that linear factor raised to increasing exponents in
the denominator.

Note of course that for linear
factors, all our numerators will be constant. So for our nonrepeated factor, π₯
minus three, we write π΄ over π₯ minus three. Then, for our repeated factor since
the exponent is three, we know we need three rational expressions. Their denominators will be π₯ plus
one in increasing exponent. So we have π΅ over π₯ plus one, πΆ
over π₯ plus one squared, and π· over π₯ plus one cubed. Letβs begin by looking at an
example that demonstrates the key process.

Find π΄ and π΅ such that two π₯
over π₯ minus three squared is equal to π΄ over π₯ minus three plus π΅ over π₯ minus
three squared.

This question is actually just
asking us to write the expression two π₯ over π₯ minus three squared in partial
fraction form. It has a repeated linear factor in
its denominator. The repeated linear factor is π₯
minus three. And we know itβs repeated because
itβs being squared. The process for performing partial
fraction decomposition in this case is as follows. We begin by checking that the order
or the degree of the polynomial in our denominator is higher than the order of the
polynomial in our numerator. If itβs not, weβll need to use
polynomial long division before splitting our remainder into partial fraction
form.

Well, if we were to distribute the
parentheses on our denominator, weβd see we have a quadratic expression. Its highest exponent is two. So the order of the polynomial on
our denominator is two. We have a linear expression on the
numerator. Its order is one. The order or degree of our
polynomial on the denominator is indeed higher than that of our numerator. So we move on to step two. Step two is to factor the
denominator if necessary. π₯ minus three squared is already
in factored form, so we move on. Then we deal with any nonrepeated
linear factors by following the usual process. There are in fact no nonrepeated
factors in our denominator, so weβre going to move on.

This is where the process differs
from dealing with nonrepeated factors. We say that if π is the exponent
of the repeated factor on the denominator, weβre going to write π rational
expressions. Each of which has that factor
raised to increasing exponents in its denominator. Note that we donβt actually
necessarily need to use increasing exponents. We could use decreasing
exponents. But by always using increasing
exponents in our denominators, we ensure that we donβt lose any of our
fractions. Our repeated factor, remember, was
π₯ minus three, and itβs raised to the power of two. So π is equal to two. And this means we need two rational
expressions when splitting it into partial fractions.

The denominator will be π₯ minus
three to increasing exponents. Well, this has actually been done
for us. We have π₯ minus three, which is
essentially π₯ minus three to the power of one. And we have π₯ minus three
squared. Note that the numerator of these
fractions will be a real constant. And thatβs because the denominator
is a linear factor. So thatβs step four completed. Whatβs the final step? We do something called multiplying
up. And what we mean by this is we add
these two fractions by creating a common denominator. And that denominator will match the
denominator of the original fraction. Once weβve done that, weβre able to
evaluate the constants, either by using substitution or equating coefficients.

So letβs see what step five looks
like. We have two π₯ over π₯ minus three
squared is equal to some constant π΄ over π₯ minus three plus some constant π΅ over
π₯ minus three squared. Note that this is sometimes written
as an identity symbol. And this is because this expression
is true for all values of π₯. In this question, though, weβre
going to keep it as an equal sign and just bear this in mind. Remember, we want to add the
fractions π΄ over π₯ minus three plus π΅ over π₯ minus three squared. And we do so by creating a common
denominator. But this common denominator will
always match the denominator on the left-hand side. Note that the denominator of this
fraction, π΅ over π₯ minus three squared, is already of this form. And so weβre actually only going to
look at this first fraction, π΄ over π₯ minus three.

To achieve a denominator of π₯
minus three squared, we need to multiply the denominator of this fraction by π₯
minus three. But of course, whatever we do to
the bottom, we have to do to the top. So weβre going to multiply both the
numerator and denominator by π₯ minus three. And so our first fraction becomes
π΄ times π₯ minus three over π₯ minus three squared. We now notice that the denominators
of our two fractions are equal. And so weβre able to add their
numerators. That gives us π΄ times π₯ minus
three plus π΅ all over π₯ minus three squared. This, of course, is still equal to
the fraction two π₯ over π₯ minus three squared. And now, we note that the
denominators of our fractions are equal.

So for the two rational functions
to be equal, their numerators must also be equal. In other words, two π₯ must be
equal to π΄ times π₯ minus three plus π΅ for all values of π₯. We have two ways to evaluate the
constants. The first way is to use
substitution. Letβs see what that looks like. We choose values of π₯ that match
the roots of our denominator. So the root of the equation π₯
minus three squared equals zero is π₯ equals three. We substitute these in in turn. So we let π₯ be equal to three, and
the left-hand side of our equation becomes two times three. The right-hand side becomes π΄
times three minus three plus π΅.

But of course, three minus three is
zero. And note that by setting π₯ equal
to three, weβve obtained an equation where the coefficient of π΄ is zero. And so we have an equation purely
in π΅. That is, π΅ is equal to six. Now we know that our identity is
true for all values of π₯. So letβs replace π΅ with six in our
original equation. We get two π₯ equals π΄ times π₯
minus three plus six. Weβre now going to choose any value
of π₯ and substitute that in. Letβs choose a simple one like π₯
equals zero. When we do, we get two times zero
equals π΄ times zero minus three plus six. Simplifying, our equation becomes
zero equals negative three π΄ plus six. And we solve for π΄ by adding three
π΄ to both sides and then dividing through by three. We find π΄ is equal to two.

So we found π΅ is equal to six and
π΄ is equal to two. Letβs very briefly look at the
other way we couldβve found these values. We go back to our earlier
equation. And we distribute the parentheses
on the right-hand side. That gives us π΄π₯ minus three π΄
plus π΅. In this method, we equate
coefficients. We compare the coefficients of our
various exponents of π₯. Letβs begin by comparing the
coefficients of π₯ to the power of one on both sides or simply π₯. On the left-hand side, we see the
coefficient of π₯ is two. On the right-hand side, we see the
coefficient of π₯ is π΄. And so we formed an equation. Itβs π΄ is equal to two.

Weβll now compare the coefficients
of π₯ to the power of zero, in other words, the constant values. On the left-hand side, there are no
constants. So the coefficient of π₯ to the
power of zero is zero. On the right-hand side, the
coefficients of π₯ to the power of zero, the constants, are negative three π΄ plus
π΅. But, of course, we just evaluated
π΄ to be two. So we get zero equals negative
three times two plus π΅. That is, zero equals negative six
plus π΅. And we solve for π΅ by adding six
to both sides. Once again, we find π΄ is equal to
two and π΅ is equal to six.

Note that both of these methods are
equally valid. In an example like this, equating
coefficients can be nice and efficient. However, with more complicated
examples, we might end up with a pair of simultaneous equations or even three
simultaneous equations in three unknowns. In that case, we would use
substitution or a mixture of both.

Letβs now look at an example which
involves more than two factors.

Determine the partial fraction
decomposition of π₯ squared plus π₯ plus one over π₯ times π₯ minus three times
π₯ plus one squared.

Here, weβre looking to determine
the partial fraction decomposition of a fraction which has a repeated linear factor
in its denominator. That repeated factor is π₯ minus
one. And we know itβs repeated because
itβs being squared. The process for performing partial
fraction decomposition in this case is as follows. The first step is to check that the
order of the polynomial in the denominator is greater than that of the
numerator. While the order or degree of the
polynomial in our denominator is full, since distributing would yield a highest
power of π₯ of four. The order of our numerator is
two. And so we can say that yes, the
order of the denominator or the degree of the denominator is greater than that of
the numerator.

Remember, if this wasnβt the case,
weβd need to use polynomial long division before splitting the remainder into
partial fraction form. Next, we factor our denominator if
necessary. Well, this is already in factored
form, and so we move on. We then look at our nonrepeated
factors. And we follow the usual
process. We essentially perform the reverse
to adding algebraic fractions. Our nonrepeated linear factors are
π₯ and π₯ minus three. So we begin the process by
splitting our fraction into π΄ over π₯ and π΅ over π₯ minus three. Remember, π΄ and π΅ are simply
constants. And this is because we have linear
factors on our denominator.

We then move on to our repeated
factor. And we say that imagine π is the
exponent of the repeated factor. Weβre going to write π rational
expressions, each of which has that factor raised to increasing exponents as its
denominator. Well, our repeated factor is π₯
plus one, and itβs raised to the power of two. In other words, π is equal to
two. So we want π rational expressions
with π₯ plus one raised to increasing exponents. Thatβs πΆ over π₯ plus one plus π·
over π₯ plus one squared. Note once again that since these
are linear denominators, their numerators are simply constants. Now we have this in the correct
form, letβs clear some space and perform the final steps.

The final step is to do something
called multiplying up. Essentially, we take our fractions
on the right-hand side and we add them by creating a common denominator. And that matches the denominator on
the left-hand side. And then we evaluate the
constants. So letβs perform that first
step. We know the denominator is going to
be π₯ times π₯ minus three times π₯ plus one squared. So what are we going to multiply
each of our fractions by? For the first fraction, weβre going
to need to multiply both the numerator and the denominator by π₯ minus three times
π₯ plus one squared. That gives us π΄ times π₯ minus
three times π₯ plus one squared over the denominator we required.

Weβll now look at our second
fraction. We see that weβll need to multiply
this by π₯ times π₯ plus one squared to achieve that denominator. And so we get a numerator of π΅π₯
times π₯ plus one squared. Weβll repeat this process for the
last two fractions. Weβre going to need to multiply
this third one by π₯ times π₯ minus three times π₯ plus one and this fourth one by
π₯ times π₯ minus three. And so the numerators of our final
two fractions are πΆπ₯ times π₯ minus three times π₯ plus one and π·π₯ times π₯
minus three. Note now that the denominators of
each of our fractions are equal. And so we can add their
numerators. And in doing so, we get the
expression shown.

Note of course that this is still
equal to our original fraction. And we might sometimes use an
identity symbol to show that this is true for all values of π₯. Weβre going to leave this as an
equal sign, but bear this in mind. Now, the whole purpose of doing
what we just did is because weβve created two fractions whose denominators are
equal. And the fractions themselves are
equal for all values of π₯. So this must mean that their
numerators must also be equal. That is, π₯ squared plus π₯ plus
one must be equal to π΄ times π₯ minus three times π₯ plus one squared plus π΅π₯
times π₯ plus one squared plus πΆπ₯ times π₯ minus three times π₯ plus one plus π·π₯
times π₯ minus three.

Now to evaluate the constants, we
have two options. We could distribute all of our
parentheses and equate coefficients. The problem is because weβre
looking to find the value of four constants, weβre going to end up having a rather
nasty set of simultaneous equations. And so instead, weβre going to
perform substitution. We look for the roots of the
denominator of our original fraction, and we substitute these in. One of the roots to the equation π₯
times π₯ minus three times π₯ plus one squared equals zero is π₯ equals zero. So letβs substitute zero into this
equation. On the left-hand side, we get zero
squared plus zero plus one, which is simply equal to one. On the right-hand side, we get π΄
times negative three times one squared. But then, π΅ times π₯ times π₯ plus
one squared becomes π΅ times zero, which is zero.

Similarly, this term becomes zero
and this term becomes zero. So the whole purpose of
substituting π₯ equals zero into this equation was to eliminate the terms involving
π΅, πΆ, and π· and form an equation purely in π΄. That equation simplifies to one
equals negative three π΄. And we solve for π΄ by dividing
through by three to get π΄ equals negative one-third. Letβs repeat this process with a
different root. Another root is π₯ equals three, so
letβs substitute that in. On the left-hand side, letting π₯
equal to three gives us 13. Then on the right, this term, this
term, and this term all contain π₯ minus three. So three minus three is zero, and
they essentially disappear. And our equation becomes 13 equals
π΅ times three times four squared or 13 equals 48π΅. Solving for π΅ gives us 13 over
48.

Thereβs one more root we can try,
that is, π₯ equals negative one. This time, letting π₯ equal
negative one eliminates this term, this term, and this term. And their equation becomes one
equals π· times negative one times negative four or one equals four π·. And so we find π· is equal to a
quarter. But how do we work out the value of
πΆ? Well, this time we can use
something called equating coefficients. Letβs imagine we were going to
distribute the parentheses on the left- and right-hand side. What would the coefficients of π₯
cubed be? Well, on the right-hand side, there
are no π₯ cubed terms, so itβs zero. If we were to distribute the
parentheses on the right-hand side, weβd get π΄ lots of π₯ cubed plus π΅ lots of π₯
cubed plus πΆ lots of π₯ cubed. So we can say that zero must be
equal to π΄ plus π΅ plus πΆ.

But of course, we evaluated π΄ to
be negative one-third and π΅ to be equal to 13 over 48. Solving this equation for πΆ, and
we find itβs equal to three over 48 or one sixteenth. All thatβs left is to replace π΄,
π΅, πΆ, and π· with their values. Weβll begin with the largest
exponent in the denominator. Thatβs this last term with π· on
it. Itβs a quarter over π₯ plus one
squared, which is one over four times π₯ plus one squared. Next, dealing with πΆ, and we get
one sixteenth over π₯ plus one, which is one over 16 times π₯ plus one. Then π΄ over π₯ becomes negative
one-third over π₯, which is negative one over three π₯. And finally, we have 13 over 48
over π₯ minus three, which we can write as shown. And weβve determined the partial
fraction decomposition of our rational function.

In this video, weβve learned that
the process for performing partial fraction decomposition with our repeated linear
factors in the denominator is roughly the same as when there are no repeated linear
factors. However, once weβve dealt with any
nonrepeated factors, we say that if π is the exponent of the repeated factor, we
write π rational expressions, each of which has that factor raised to increasing
exponents as their denominators. We then multiply up and use
substitution or equating coefficients to calculate the value of the constants.