Welcome to Part One of the lesson,
on Partial Fraction Decomposition.
Partial Fraction Decomposition is a process used
to rewrite rational functions
as a sum or difference of simpler rational functions
which in calculus are easier to integrate.
We know from algebra, we often add or subtract,
rational expressions by obtaining a common denominator
and then adding or subtracting the numerators
to write the sum or difference
as a single rational expression.
However, partial fraction decomposition,
is just the opposite of this process.
For example, in calculus, we may have a difficult time,
integrating a specific rational function
because it does not fit a specific integration formula.
However, we can sometimes write the rational function
as a sum or difference of simpler rational functions
which are easier to integrate
which is why the process
of partial fraction decomposition can be very useful.
However, before performing partial fraction decomposition,
if the degree of the numerator is greater than,
the degree of the denominator,
then we must divide the denominator,
into the numerator to obtain a polynomial,
plus a rational function with a degree of the numerator
or P sub one of x is less than
the degree of the denominator, or in this case Q of x.
Once we have the rational function in the correct form,
the first step completely factor the denominator,
into prime linear or quadratic factors.
Step two, for linear factors, each distinct linear factor,
in the form ax plus b must include a term,
in the following form where we have a single,
variable in the numerator
and the variables in the numerator are not repeated.
For each repeated linear factor in the form,
the quantity ax plus b raised to the power n,
we must include n terms of the following form.
Again notice how the variables in the numerator,
are never repeated.
For quadratic factors each distinct quadratic factor,
in the form ax squared plus bx plus c,
must include a term in the following form.
But notice now the numerator is a linear factor.
For repeated factors in quadratic form,
raised to the power of n we must again include n terms
of the following form.
And once again, notice how the variables in the numerator,
are never repeated.
For step three once we have the partial fractions,
we multiply each fraction by the LCD.
And the result is what's called the basic equation
which will then solve for A, B, C and so on.
So let's perform partial fraction decomposition,
on this rational expression,
which we should remember from the first slide,
should give us one divided by the quantity x plus three,
minus one divided by the quantity x plus five
because we know the difference,
is this rational expression here.
Let's just verify we get the same result
by performing partial fraction decomposition.
So again, the first step is to recognize,
the degree of the numerator is less than,
the degree of the denominator and therefore we can go ahead
and proceed to partial fraction decomposition
and factor the denominator.
And the denominator factors in a two linear factors.
The factors of x squared are x and x,
the factors of 15 and eight,
are positive three and positive five,
which gives us x plus three times x plus five.
And now because we have two distinct,
linear factors in the denominator,
this rational expression is equal to A,
divided by the linear factor of x plus three,
plus B divided by the linear factor of x plus five.
Now if we had A over x plus five and B over x plus three,
that would be just fine as well.
The next step is to clear the fractions from the equation
by multiplying both sides of the equation
or each rational expression by the least common denominator
which is x plus three times x plus five.
So let's go ahead and set that up.
If it's helpful, they can write the factors of x plus three
and x plus five as a fraction with the denominator of one.
And now before multiplying,
we will simplify out all the common factors,
between the numerators and denominators.
And we see we have a common factor of x plus three
as well as a common factor of x plus five,
we notice we have two on the left,
we have two equals on the right side,
a factor of x plus three simplifies out here,
leaving us with A times the quantity x plus five
and then we have plus,
here the common factor of x plus five simplifies out.
Leaving us with B times the quantity x plus three.
So this is our basic equation
which we now need to solve for A and B.
Let's take this to the next slide.
This equation is true for all values of x.
However, if we select convenient values of x,
it makes it much easier to solve for A and B.
Notice how if we let x equal negative five,
x plus five would be zero.
If we let x equal negative three,
x plus three would be equal to zero.
So let's begin by letting x equal negative five.
If x is negative five, we have the equation two,
equals A times the quantity x plus five
but if we substitute negative five for x,
we have negative five plus five which is zero,
A times zero is zero.
And then we have plus B times the quantity negative five,
plus three which is B times negative two or negative two B.
Solving for B here, we divide both sides by negative two,
which gives us B equals negative one.
And that is all for A, let's let x equal negative three.
When x is negative three,
we have the equation two equals,
A times the quantity of negative three plus five,
which is A times two or two A.
And then again, when x is negative three,
x plus three is zero, B times zero is zero.
Which gives us the equation two equals two A,
dividing both sides by two we have A equals one.
And now we have the information we need
to determine the partial fractions
of the original rational expression which was two,
divided by the product of the quantity x plus three
and x plus five which remember we said was equal to A,
divided by the quantity x plus three plus B,
divided by the quantity x plus five.
And now that we have A and B
and that we know A and B,
we know the given rational expression,
is equal to A which is one,
divided by the quantity x plus three
and then we have plus B divided by the quantity x plus five
and since B is negative one,
we can write that as minus one,
divided by the quantity x plus five.
So this is the partial fraction decomposition
of two divided by the product of the quantity x plus three
and x plus five which if we go back to the first slide,
is verified from the difference shown here.
And again, the point of this is,
in calculus if we have this integral,
we can now find the difference of these two integrals
which is much easier to do,
using this basic integration formula.
Let's look at one more example before we go.
Determine the partial fraction decomposition
of this given rational expression.
So again, first, notice how the degree of the numerator,
is less than the degree of the denominator.
So the next step is to factor the denominator,
the first step is to factor out the greatest common factor
which is x, if we factor out x,
we're left with the quantity x squared plus two x plus one
and now we keep factoring,
x squared plus two x plus one will factor
to two binomial factors.
The factors of x squared are x and x,
the factors of one and two are one on one
and therefore we have two factors of x plus one.
So let's write this as.
So let's write this as the quantity three x squared,
plus three x plus one.
And all this is divided by the product of x
and the quantity x plus one squared.
And now set up the partial fractions.
x is a linear factor and therefore we have A,
divided by x and then plus we have the repeated factor
of x plus one so we have B,
divided by the quantity x plus one.
And then because x plus one is a repeated factor,
we also include plus C divided by two factors of x plus one,
or the quantity x plus one squared.
And now the next step,
we multiply every fraction in the equation
by the least common denominator
which is x times the quantity x plus one squared.
So let's go ahead and set this up.
Let's write x times the quantity x plus one squared,
as a fraction with the denominator of one.
And now we simplify before multiplying.
Notice here we have x divided by x
as well as the quantity x plus one squared,
divided by itself, leaving us with just three x squared,
plus three x plus one equals on the right side,
here only a factor of x simplifies out,
leaving us with A times the quantity x plus one squared.
Then we have plus, here only one factor of x plus one,
simplifies out so this simplifies,
the one that simplifies to one factor of x plus one,
leaving us with Bx times the quantity x plus one.
And then here two factors of x plus one simplify out,
leaving us with C times x or CX.
So this is our basic equation which you now need to solve,
for A, B and C and let's do this on the next slide.
And now this equation is true for all values of x
but again, we'll start giving it it values of x
to make it easy to solve for A, B and C.
We'll let x equals zero and x equal negative one.
So first let x equal negative one.
If x is negative one here we would have three times,
the square of negative one plus three times,
negative one plus one equals,
if x is negative one, x plus one is zero here
as well as here leaving us with just C times,
negative one or negative C,
so here we have three plus negative three plus one,
that's one, one equals negative C
and then for C is equal to negative one.
And now if we let x equals zero.
When x is zero, we have zero plus zero plus one
which is one equals on the right side.
When x is zero, this would be A times the square of one
which is just A times one or A.
This would be zero and this would be zero
and therefore A is equal to one.
So now we know the values of A and C,
we still need to find the value of B though.
So we need to select another value of x,
let's try x equals one.
When x is one, on the right side we have three,
times one squared plus three times one plus one
that's seven equals when x is one here we have,
the square of two which is four.
Four times A is four A but we know A is one
so we have four times one plus,
when x is one we have B times one times two that's two B
and then we have plus C times one but C is negative one
which gives us just minus one.
Simplify and we have seven equals four plus two B,
minus one that's three plus two B.
Subtracting three on both sides gives us four,
equals two B dividing both sides by two,
we have B equals two.
Now that we know the values of A, B and C.
We now know the partial fraction decomposition,
for the visual rational expression which is this line here.
Let's go ahead and write this out.
The given rational expression is equal to A,
divided by x where is one which gives us one divided by x,
plus B divided by the quantity x plus one
which gives us plus two divided by the quantity x plus one
and then plus C divided by the quantity
of x plus one squared and since C is negative one,
this gives us minus one divided by the quantity,
x plus one squared.
This is the partial fraction decomposition
of the given rational expression.
I hope you found this helpful.