>> Welcome to the Cypress College Math Review
on Partial Fraction Decomposition.
Partial fraction decomposition is a method of breaking
up a complicated rational expression
into the sum of simpler ones.
First factor the denominator of the rational expression.
Based on the chart below determine what terms will be
in your decomposition.
When you factor your denominator if you end up with a factor
that looks like this, that's the first power of a linear,
linear mx + b. So 3x + 1 is the first power of a linear.
You then will end up with the term a over 3x + 1
in your decomposition.
If instead you ended up with the square of 3x + 1,
that's the second power of a linear.
Then you would have a over 3x + 1 + b over the square of 3x + 1.
You would have 2 terms in your decomposition
because it's the second power of a linear.
Other examples of first power of linears are and, other examples
of second power of linear are and.
Now let's look at an irreducible quadratic.
An irreducible quadratic is one with no real 0s.
For example, x squared + 4.
If you set that equal to 0, you don't get a real number.
If that is a factor of your denominator, then instead
of putting a over that in your decomposition,
you will put ax + b over that.
If you have the second power of that, then you will put ax + b
over it and then you will also have a second term
where you will put cx + d over that squared.
This is how you set up your decompositions.
So the first step is to factor the denominator.
Each of these factors are the first power of a linear.
So we end up with a over x + 2 and b over x - 1.
The next step is to multiply by the original denominator.
Over on the right hand side when I multiply by the denominator,
it looks like I am cross multiplying.
There are 2 different methods
for finishing a problem at this point.
You can either equate coefficients or substitute
in values for x. On the first problem,
I'm going to equate coefficients.
On the other problems, I will substitute in values for x.
So to equate coefficients, I distribute
and basically I am adding like terms
and factoring out the powers of x.
So on the left, I have 3x, on the right I have (a + b)x.
So this is supposed to be true for all x, therefore,
3 is supposed to equal a+b.
On the left hand side, my constant is 0,
on the right hand side, my constant is negative a + 2b.
Therefore, 0 is equal to negative a + 2b
and then you solve the system in any way you would like to.
So b is equal to 1.
So a is equal to 2.
So the answer to the problem is 2 over x + 2 + 1 over x - 1.
Before breaking up a rational expression using partial
fraction decomposition you must first make sure the expression
is proper.
A proper rational expression is one in which the degree
of the numerator is less than the degree of the denominator.
If the rational expression is improper, then divide first.
Use partial fraction decomposition to break
up the rational expression that you're left with.
On this problem, you notice that the degree of the numerator is 3
and the degree of the denominator is 2.
So we have to divide.
In fact if the degrees were the same,
you would also have to divide.
So we do our long division.
We subtract we get negative 2x squared + 4x - 3.
That goes in negative 2 times
so we get negative 2x squared - 6x + 8.
We subtract and get 10x - 11.
So, our quotient is x - 2, our remainder is 10x -
11 and our divisor is x squared + 3x -4.
We still need to do a partial fraction decomposition
on this fraction here.
So eventually we'll put our answer down here.
So let's factor our denominator.
Both first power of linear.
Multiply by the denominator, the original denominator.
It looks like we're cross multiplying.
This time I'm going to substitute
in our critical values of 1 and negative 4.
That will make this term disappear.
So if we let x be 1 on the left hand side, we get this.
On the right hand side,
the first term disappears and we get 5b.
So we get negative 1 is equal to 5b.
So b is equal to negative 1/5.
Now we'll let x be the other critical value, negative 4.
This time this term will disappear
and we'll have negative 5a on the right hand side.
So a will be 51/5.
So the partial fraction decomposition gives us a result
of 51/5 over x + 4 - 1/5 over x - 1,
but normally you don't write fractions inside fractions;
that's a complex fraction.
So let's multiply 5 over 5 to those fractions
as we put our answer over on the left hand side here
with our X -2.
So we have 51 on the top and we have 5 times the quantity X + 4
and then we'll change that to subtraction and then 1
over 5 times the quantity x - 1 and there's our answer.
All right here we go it's all factored for us and ready to go.
So we have a over x + 1.
The second factor is an irreducible quadratic.
So instead of just b over that factor we will have bx + c
multiply by the original denominator.
Watch your parentheses over here.
So let's let x be negative 1.
So on the left hand side, we get 1 + 4 + 7.
On the right hand side, this term goes away
but over here we have a times the quantity 1 + 2 + 3
which gives me 12 is equal to 6a so a is 2.
And now we don't have any more critical values
so we just pick numbers we like.
So let x be 0.
So on the left hand side I got 7,
on the right hand side I got 3a + c. I already know that a is 2,
which gave me that c is 1 and then let's let x be say 1.
So on the left hand side, I get 1 - 4 + 7 that's equal
to a times 1 - 2 + 3 + the quantity b + c times 2.
You'll notice that I'm substituting in for x but not
for a, b and c. I usually do it for x and then I do it for a,
b and c on the next step, but whatever works for you.
So on the left hand side, I get 4 is equal to 4 + 2b + 2
so negative 2 is equal to 2b which gives us
that b is equal to negative 1.
Answer a, which is 2 over x +1,
+the quantity let's see what have we go?
Negative 1 for b so that's our coefficient
for x and then c is 1.
So negative x + 1 is the numerator in the second fraction
and our denominator was x squared - 2x + 3.
The first factor is the first power of a linear.
The second is the second power of a linear.
So we not only have b over x + 1, but we also have c
over the square of x + 1.
It is not cx + d. Now multiply by the original denominator
and we end up with x - 3 on the left hand side.
On the right hand side, we have a times the square
of x + 1 + b times x + 2 x + 1 + c times x + 2.
First we'll let x be negative 1.
In this case, we get negative 4 on the left, on the right,
we simply get c. So we have our first value.
Next let x be negative 2,
on the left hand side we get negative 5,
on the right hand side we simply get a. So a is negative 5.
Finally, we run out of critical values
so we'll let x be some other value say 0.
On the left hand side, we get negative 3,
on the right hand side substituting in 0
for x we get a + 2b + 2c.
Now substituting in the values we know for a
and c we have negative 5 + 2b -8.
So we have 10 is equal to 2b which tells us that b is 5.
That gives us our answer of negative 5 over x + 2 + 5
over x + 1 - 4 over the square of x + 1.
Here's some more problems that you should practice on your own.
After you have done the problems,
check with the answers below.
Here are the answers to the practice problems.