Ex: Partial Fraction Decomposition – Degree 2 / Degree 3
Ex: Partial Fraction Decomposition – Degree 2 / Degree 3


We want to find the partial fraction decomposition of the given rational expression, which means we wants to write
the given rational expression as a sum or difference of
simpler rational expressions. The first step is to make sure the denominator is in factored form and because our denominator has four terms let’s try to factor by grouping. So if we want to factor x to the third plus two x squared plus
nine x plus 18 by grouping, we factor out the greatest common factor from the first two terms, then the greatest common
factor from the last two terms and then hopefully we have
a common binomial factor. So, the greatest common
factor of the first two terms would be x squared, so we’d have x squared times the quantity x plus two. We notice that nine is
the greatest common factor of the last two terms. Because we’re factoring
out a positive nine, we’ll write plus nine times
the quantity x plus two. You notice how these two products do have a common binomial
factor of x plus two. If we factored the x plus two out you’ll notice how we’d be
left with x squared plus nine. And now we have the factored
form of our denominator so the given rational expression is equivalent to the same numerator. And now the denominator in factored form would be the quantity x plus two times the quantity x squared plus nine. So notice how we have a linear factor and a quadratic factor,
which help us determine the partial fraction decomposition. Again, the given rational expression with the factored
denominator would be equal to the sum of two rational expressions where the first denominator would be the linear factor of x plus two and the second fraction
would have a denominator of x squared plus nine. And now for the fraction
with a linear denominator, the numerator would be a
constant which we’ll call A. And now for the second
rational expression here that has a degree two denominator
or a quadratic denominator we would have a linear numerator
in the form of B x plus C. Now the next step will be to clear the fractions from our equation, which will give us what’s
called the principal equation. So we’ll multiply both
sides of the equation by the least common denominator, which would be these two factors. Notice the left side of the equation simplifies nicely to seven x squared plus 11 x plus 46 and the right side of this first product, notice how the factor of x plus two simplifies out or divides out leaving us with A times the
quantity x squared plus nine. And for the second product here, the factor of x squared
plus nine simplifies out leaving us with plus
the quantity B x plus C times the quantity x plus two. And now to find the values of A, B, and C, we’ll equate the coefficients. So for the next step, we’ll
multiply out the right side. So we’d have A x squared plus nine A plus B x squared plus two B x plus C x plus two C. Now we’ll group the x
squared terms together, the x terms, and the constants. So we have A x squared plus B x squared and then we have two x terms. We have two B x plus C x and the constant terms
would be nine A and two C, so plus nine A plus two C. Now let’s factor the x squared
out of these two terms, the x out of these two terms, and then we’ll just group
the constants together. So, let’s go ahead and put the
x squared factor on the right so we’d have the quantity
A plus B times x squared plus, let’s go ahead and factor out the x and put the factor of x on the right so we would have two B plus C and then we just have plus the
quantity nine A plus two C. And now we equate the coefficients to form a system of equations, meaning the coefficients
of the x squared terms must be the same, so
seven must equal A plus B, 11 must be equal to two B plus C, and finally, 46 must
equal nine A plus two C. So again, we can say that
A plus B must equal seven, two B plus C must equal 11, and nine A plus two C must equal 46. So now we’ll have to solve
the system of equations to find the value of A, B, and C. Let’s go ahead and do
this on the next slide. Let’s use elimination and eliminate B from these first two equations. So notice how if these was negative two B, we could add the two equations together to eliminate the B term. So let’s go ahead and multiply this first
equation by negative two and we’ll leave the
second equation the same. I’m going to go ahead
and put the B term first so we can write this as negative two B minus two A would be equal to negative 14 and the second equation stays the same. We have two B plus C equals 11. So notice how this gives us an equation that contains A and C and then
we can use the third equation to solve for A and C. So here we have negative two A
plus C equals negative three. So now we’ll use the third equation, nine A plus two C equals 46 and this equation that we just formed, negative two A plus C
equals negative three and we’ll solve this system for A and C. Now we’ll before and back
substitution to find B. To eliminate C, we would
want this term here to be a negative two C, so we’ll multiply the second
equation by negative two and leave the first equation the same. So nine A plus two C equals 46. Here we’d have four A minus
two C equals positive six so the C’s simplify out. We have 13 A equals 52. Dividing both sides by 13, notice we found that A equals four. So A equals four and now we’ll perform back
substitution to find B and C. We’ll notice here if A is four, we know A plus B is seven, so B has to be three. And if B is three, two
times B would be six. Six plus 5 is 11, so C equals five. These are the values that we needed to find the partial
fraction decomposition. We’ll now substitute A, B, and C here, here, and here, and this will be our partial
fraction decomposition. The partial fraction decomposition of the given rational
expression, because A is four, would be four divided by
the quantity x plus two plus, because B is three and C is five, we’d have three x plus five divided by the quantity
x squared plus nine. This is our partial
fraction decomposition. I hope you found this helpful.

3 thoughts on “Ex: Partial Fraction Decomposition – Degree 2 / Degree 3”

  1. Catch 22 says:

    Finding the constants would be easier if you plug in x=-2 at 3:05 and solved.

  2. GashPlague says:

    Is the "Bx+C" numerator form also used for denominators with x to powers higher than 2?  For example: If the denominator was "(x + 2)(x^3 + 9)", would you still use "Bx+C" as the numerator in your decomposition, or would you use a modified form?

  3. Make or Break TV says:

    thank you !

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