Welcome to the first of several examples
of finding the partial fraction decomposition
of a given rational expression.
Partial fraction decomposition is a method
for writing a given rational expression
as a sum or difference of rational expressions
that are simpler than the given expression.
And this is often helpful when integrating in calculus.
Before we start, we need to make sure
that the degree of the numerator
is less than the degree of the denominator.
For example, here we have 12,
which is degree zero,
and in the denominator we have degree two,
so we have satisfied the first condition,
so we can go ahead
and perform partial fraction decomposition.
If the degree of the numerator
was not less than the degree of the denominator,
we'd have to start by performing long division.
So the first step will be to
factor the denominator completely,
so we would have 12 divided by
x squared minus nine is a difference of squares,
so we'd have a factor of x plus three
and a factor of x minus three.
Notice in this case,
our denominator contains two distinct linear factors,
which means we'll have two fractions
in our partial fractions,
where the first fraction
will have a denominator of x plus three,
and the second fraction
will have a denominator of x minus three.
Now if we had repeated linear factors,
or quadratic factors in the denominator,
it would affect how we set these fractions up,
which we'll see in future examples.
And again, because we have linear factors
in our denominator,
that means our numerators will be constants
which we don't know the values of right now,
so we'll use the variable A for this numerator,
and B for this numerator.
So we're using a single variable for the numerators
because we have linear factors in the denominator.
And now if we can find the value of A and B,
we'll have our partial fraction decomposition.
So to solve this equation,
we're going to clear the fractions
by multiplying both sides of the equation
by the least common denominator,
which would be the quantity x plus three
times the quantity x minus three.
Notice on the left side when we multiply,
x plus three over x plus three simplifies to one,
and so does x minus three over x minus three,
so we're left with twelve
equals
and on the right side, for this first product,
x plus three over x plus three would simplify out.
Leaving us with A times the quantity x minus three,
and then for the second product,
notice how x minus three over x minus three
simplifies to one, so we'd have
B times the quantity x plus three.
This equation here is called the basic equation,
which we're going to solve to find the value of A
and the value of B.
And there are a couple ways to solve this for A and B.
One way is to multiply out the right side,
combine like terms,
and then set the coefficients of like terms
on the left side of the equation
and the right side of the equation
equal to each other,
to form a system of equations.
Or, which we'll do in this example,
we can select convenient values of x
and then perform substitution
in order to find the values of A and B.
When I say convenient values of x,
we're going to pick values of x
that make the factor x minus three
or the factor x plus three equal to zero.
For example, if we select x equals three,
and perform substitution,
that would give us the equation 12 equals,
if x is equal to three we would have A times zero
which is equal to zero,
so this is eliminated.
So we have B times quantity three plus three
which would be six B,
divide both sides by six,
we know that B must equal positive two.
Another convenient value of x would be the value
that makes x plus three equal to zero,
which would be x equals negative three.
So we'll substitute negative three for x,
which would give us the equation 12 equals,
this would be A times the quantity
negative three minus three, or negative six A,
and here, negative three plus three is zero,
B times zero is zero,
giving us this one step equation.
Divide both sides by negative six.
And so we have A equals negative two.
And now we'll substitute the values of A and B
into our fractions here on the right,
which will be our partial fraction decomposition.
Notice that A would be negative two
and B would be positive two,
so we could leave it like that, but it'd be more common
to write the positive fraction first.
So we found that 12 divided by the quantity x plus three
times the quantity x minus three is equal to,
again, we're going to put the second fraction first,
B is positive two,
so we have two divided by the quantity x minus three,
and then because A is negative two,
instead of writing plus negative two,
we'll write just minus two,
divided by the quantity x plus three.
This would be the partial fraction decomposition
of our rational expression.
Meaning if we subtracted these two fractions
by obtaining a common denominator,
we would end up with the fraction on the left.
Okay, we'll look at several more examples
in the next few videos.
I hope this was helpful.