Welcome to a second video on
partial fraction decomposition.
In this video we'll take a look at having quadratic factors.
This video does assume you've already watched part one,
so we're going to go ahead and go quickly
into the examples after a quick review.
Before we preform partial fraction decomposition
if the degree of the numerator is greater than
the degree of the denominator we do need to
divide the denominator into the numerator
to obtain a polynomial and a rational function.
Where the numerator is less than
the degree of the denominator.
Once we have the rational function in the correct form
we're going to factor the denominator completely
into linear and quadratic factors.
Get the linear factor we'll have
single variables in the numerators.
And if a linear factor is repeated we do have to
write an extra fraction for each repeated linear factor.
For quadratic factors we'll have a
linear numerator in the form of BX plus C.
And again if the factor is repeated we
do have to have additional fractions
for each repeated factor with a different numerator.
Once we have this, once we have the partial fractions
we're going to multiply each fraction by the LCD
and solve the resulting basic equation.
The process is a little more involved with quadratic factors
We have to expand the basic equation
and then combine like terms based on the power of x
to set up and solve a system of equations.
Let's go ahead and take a look at our first example.
To save a little bit of time,
the denominator is already in factored form.
And this is not factored because there's no factors
of positive three that add to negative two.
So we have a linear factor and a quadratic factor.
So for the linear factor we'll have A over x minus two.
Plus, for the quadratic factor we'll have Bx plus C.
All over x-squared minus two x plus three.
Now we'll multiply every fraction by the LCD.
Which is x minus two times x-squared minus two x plus three.
Notice on the left side both factors simplify out,
we're left with seven x-squared minus 13 x plus 13,
must equal, here only the factor of
x minus two simplifies out.
So we have A times x-squared minus two x plus three,
plus here, the quadratic factor simplifies out.
So we have Bx plus C times x minus two.
Now because we have a quadratic factor
we are going to have to multiply all this out.
But before we do that, if we use the zero
of the linear factor we should
be able to determine either A, B, or C.
Let's go ahead and try that.
So if x equals two, if we sub a two in here
we'd have four times seven that's 28.
Minus 26 plus 13 that'll give us 15.
Equals, again if A is two, we'd have four
minus four plus three, that'll give us three times A.
And then if x is two, this would be
zero times anything would still be zero.
So we have 15 equals three A so
we know that A is equal to five.
But to determine the values of B and C
we are going to have to multiply all this out.
And then collect the like terms based upon the power of x.
So we're going to have seven x-squared minus 13 x plus 13
equals A x-squared minus two A x plus three A
plus B x-squared minus two B x plus C x and then minus two C
Now we're going to combine the like terms
based on the power of x on the right side.
So here are the x-squared terms.
Here are the x terms.
And here are the constant terms.
So we'll have A plus B x-squared.
Plus negative two A minus two B plus C times x.
And then for the constant terms we just have
three A minus two C.
Now we're going to match up the coefficients meaning,
A plus B must equal seven.
All of this must equal negative 13.
And all of this must equal positive 13.
Let's start with A plus B must equal seven.
Remember we already found that A was equal to five,
so if A is five we know that B is going to equal two.
So now we just need to find C.
Let's take a look at the constant expression here.
We know that three A minus two C must equal 13.
But again we know that A is equal to five,
So this would be 15 minus two C equals 13.
We subtract 15 on both sides we'll have
negative two C equals negative two.
So now we know that C is equal to positive one.
So now we know the values of A, B, and C.
So now we can rewrite this rational expression
as a sum of these two fractions.
So this rational expression can be written as,
A over x minus two which is five over x minus two
plus Bx plus C over x-squared minus two x plus three.
So we'd have two x plus one over x-squared
minus two x plus three.
This is the partial fraction decomposition
of this given rational expression.
Let's see if we have time for one more.
The degree of the numerator is less than
the degree of the denominator so lets
go ahead and factor the denominator.
This will have two quadratic factors.
The factors of two that add to three,
that would be plus two and plus one.
And since these are quadratic factors
the first one will be Ax plus B over x-squared plus two.
And the second fraction will be
Cx plus D all over x-squared plus one.
Now we're going to multiply every fraction by the LCD
which will be the quantity x-squared plus two
times the quantity x-squared plus one.
On the left side these factors will simplify out
we'd be left with x-cubed plus x-squared plus x plus two
must equal here only the factor of
x-squared plus two would simplify out.
So we'd have Ax plus B times the quantity x-squared plus one
and on this fraction the x plus one would simplify out
so we'd have Cx plus D times x-squared plus two.
Since we don't have any linear factors here
we are going to have to multiply this out
on the right side and then combine the like terms.
So we'd have A x-cubed plus Ax plus B x-squared
plus B plus C x-cubed plus two Cx
plus D x-squared and then plus two D.
Let's go ahead and color code the like terms.
So let's go ahead and combine our like terms now
in terms of the powers of x.
So we'd have A plus C x-cubed plus B plus D x-squared
plus A plus two Cx and then lastly plus B plus two D.
LEt's go ahead and include the coefficients here,
we have one, one, one, and then two.
We have A plus C equals one.
B plus D equals one.
A plus two C equals one.
And B plus two D equals two.
LEt's go ahead and call this equation one,
equation two, equation three, and equation four.
Let's go ahead and take equation three
and subtract equation one.
So we'd have A minus A that would be zero.
Two C minus C would be C.
And then one minus one would be zero.
So C is zero.
Well next if C is zero than A plus zero
must equal one so A is equal to one.
Now let's take a look at equation four and equation two.
LEt's take equation four and subtract equation two.
So we have B minus B that would be zero.
Two D minus D that would be D.
And then two minus one would be one. So D is equal to one.
Now looking at equation two if D
is equal to one B must equal zero.
So we have all the information we need.
We just need to fill in the values of A, B, C, and D
into these fractions here.
Let's go ahead and do that on the next screen.
Okay, so here's the original equation that we had.
Here are the fractions we want to find.
And now we have the values of A, B, C, and D.
So this rational expression is equal to
Ax plus B over x-squared plus two.
Well A is one and B is zero so we'd have
x over x-squared plus two plus,
and for this fraction, C is zero and D is one.
So we'd have one over x-squared plus one.
And this would be the partial fraction decomposition
of this given rational expression.
Okay, hope you found this second video helpful.
Next we'll take a look at combining
this with integrating rational functions.
Thank-you for watching.