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.

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

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?

thank you !