Welcome to the first of two videos on how to perform partial fraction decomposition.
This video will only focus on the linear factors.
Partial fraction decomposition is a process used to rewrite
rational functions,
as a sum or difference of rational functions
that are easier to integrate.
In algebra class,
we often took rational expressions and combine them,
by obtaining a common denominator,
to rewrite the sum or different as a single rational expression.
However, in calculus,
If we have a rational function this difficult to integrate
we can often,
write it as a sum or difference
of simpler a rational functions
making it easier to integrate.
Let's take a look at how we're going to do this.
First,
If the degree at the numerator is greater than the degree of the denominator,
we want to divided the denominator into the numerator
to obtain
a polynomial
and a rational function,
where the degree of the numerator is less
then the degree of the denominator.
Once we have in this form,
we wanna completely factor the denominator
into prime linear or quadratic factors.
For each linear factor,
will have nominator of
a single variable A
over the near factor.
If the linear factor repeats,
we would have to include additional fractions,
picking up each
additional linear factor over a new numerator.
until it become comminuted for all of the repeatedly linear factors.
And, if we have a quadratic factor,
and if we have a quadratic factor,
notice our numerator going to be in the form of,
Bx plus C.
So, we have a binomial linear factor,
over the quadratic factor.
And again, if it's repeated we would have to include,
additional fractions
with different name numerators.
And then lastly, once we have the partial fractions
we're gonna multiply,
each fraction by the LCD,
and the result is called the basic equation.
and then we gonna solve the basic equation by determining the values of A, B, C, and so on.
So, that we can rewrite the rational function,
as a sum or difference of simpler rational functions.
So, lets go and look at an example.
We want to rewrite this,
as a sum or different of simpler rational expressions.
First thing we should check is that the numerator has a
degree less then a denominator, which it does.
So, lets go and factor this denominator
take away X plus five times X minus three
this we have two distinct linear factors.
So, that tells us that we gonna set this up as
A over one of linear factors. Lets call it x plus five,
plus
B,
over the other linear factor of X minus three.
Now, we'll multiply every faction by the LCD, which is X plus five times X minus three.
Notice on the left side,
all these factors simplify out.
So, we left with two
equals,
here the x plus five simplify out. We are left with A times X minus three
plus,
here the X minus three simplify out,
and we are left with B times X plus five.
This is our basic equation
let's go and take this over the next screen and continue.
Now we need to solve this basic equation by determining the values of A and B.
It's a pretty straight forward process, when we have linear factors.
This equation is true for all values of X.
Which means we can pick any value of X we want, to determine the values of A and B.
Notice here, if we let X equals three,
this factor here would be equal to zero.
let's go and do that.
Let x equal three
then we'll have two equals.
A times three minus three. that would be zero,
plus
B times
three plus five, that would be eight.
dividing both sides by eight.
this tells us that B is equal to one-fourth.
Now, looking at this binomial factor,
Let X equal negative five.
Remembers is true for any value X
So we have linear factors. If we picked the zeros of the linear factors,
it'll make finding the other variables very straight forward.
So X is negative five. We would have,
two equals, it should be A times negative eight
plus
B time zero. So, divide both sides by negative eight.
We have A equals negative one-fourth.
So going back to the previous screen, just for a moment,
Notice that we just found the value of A, which is the numerator of
the fraction with denominator of x plus five,
and B is a numerator of the fraction
with the denominator of X minus three.
I copied it down here in red
so we've just discovered is that,
If we decompose
this fraction it would be equal to
well A equals to negative one-fourth.
We'd have negative one over four times X plus five
plus, B which is positive one-fourth. So, would have one over
four times X minus three.
This is the fraction decomposition of
two
over X square plus two x minus fifteen.
One thing you might have noticed already, if you go back to the first screen
that we said that this was equal to
one over x plus five
plus one over x minus three,
and this is different from we just found
doesn t what we found was wrong, it's just a different form
Just to verify, I'm gonna show if we do combine these will get
two over X squared plus two x minus fifteen.
We want to combine these. We would have to get a common denominator that would be
four times X plus five times X minus three
this fractions is missing the X minus three
this fraction is missing the X plus five
so we have four
times X plus five times X minus three
and we would have
negative x plus three
and then X plus five
you can see we have eight
over
four times X plus five
times X minus three
and then lastly this would simplify here,
to a one into a two
which does match
the original
rational expression.
Let's go and try one more.
In the first step, is to recognize that degree of the numerator is less then the degree of denominator.
So, now we go ahead and factor the denominator completely.
we have a common factor X
this is a perfect square trinomial. So, will have X times X plus one
time other factor X plus one
So, we have, three X squared
plus three X plus one over
X times X plus one squared
is gonna equal; notice, all the factors are linear
So, we'll have a single variable in the numerator, we will have A over X
plus
B over X plus one.
And since, this factor is repeated. We need in another
fraction where we have C over the quantity X plus one to the second
Now, we gonna multiply,
everything in the equation by X times X plus one squared
Notice, on the left,
everything simplifies out.
we are left with three X squared,
plus three X plus one,
would equal
notice here, only factor of x simplifies out. So, we are left with
A times the quantity X plus one squared
plus,
here notice, only one factor of x plus one simplifies out.
So, have B times X times X plus one
plus.
Here two factors of x plus one simplifies out; we are left with C x.
Okay, this is our basic equation
Now we need to determine the value of A, B, and C,
by selecting values of X.
We can select any value of X that we wish but selecting the zeros,
of these leaner factors
should make it lot easier
let say, X equal zero.
If X equals zero
on the left we have zero plus zero plus once, we have one equals,
Here we'd have one squared times A, that would be A
well here's an X that would be zero,
and here is an X and that would be zero.
so we know that A is equal to one
Now, lets use x equals negative one
X equals negative one,
here we 'd have three
plus negative three that's zero plus ones. So, we have one
equals
If X is negative one, this should be zero
If X is negative one this would be zero,
and if X is negative one, this should be negative C.
So, one equals negative C, we know that the C equals negative one.
And now we need to determine the value B.
So we need to select another value of X
but we can also use the fact that we know A is one, and C is negative one.
so let's say X equal one.
If X is one on the left, we would have
three plus three plus one that be seven equals
Here we would have one plus one that would be two square that would be four,
times A. But now, we know A is one, that would be four.
plus
when X is one, here we have one plus one that would be two,
times one
times B, that would be two B.
and then, we have C times X, But know, C is negative one
So, we have negative one times one that would be negative one.
looks like we have, four munis one that would be three, so taking three on both side, so we have
four equals two B
So looks like B equals two.
Okay, we have everything we need.
we gonna go ahead and rewrite this
partial fraction decomposition here
with the values of A, B, and C as the numerator.
Let's go and do that on the next screen
Okay, here's how we started. We had this rational expression,
we would wrote it as a sum of
these three fractions and then we determined the values of A, B, C
from the basic equation.
So, we have found that this rational expression is equal to
one over x
plus
two over X plus one
and since C is negative will have minus,
one over X plus one squared.
this is the partial fraction decomposition of this rational expression.
Next, we'll take a look at how to perform partial fraction decomposition
when we have a quadratic factors
Hope you found this helpful.