Partial Fraction Decomposition

Partial fractions (in a "special case")

Partial fractions is a way of taking a rational function (a function of the form $p(x)/q(x)$, where $p, q$ are both polynomials) and "splitting it up" into fractions. An example is $$\frac{3x-2}{x^2-4} = \frac{1}{x-2} + \frac{2}{x+2}$$

Let's find a way to do this. Let's first do it when the denominator is a quadratic, and use the function $\frac{3x+1}{x^2-x-6}$ as an example.

In the spirit of the example above, let's try to write $$\frac{3x+1}{x^2-x-6} = \frac{c}{x-a} + \frac{d}{x-b}$$

Why are they $-2, 3$? Because if you add $\frac{c}{x-a} + \frac{d}{x-b}$ together using a common denominator, that common denominator will be $(x-a)(x-b)$. Therefore, this must equal the denominator on the left hand side. So $(x-a)(x-b) = x^2-x-6$, which factors nicely as $(x+2)(x-3)$. A more general statement would be that $a, b$ are the roots of the polynomial $x^2-x-6$.

So now we are working with the equation $$\frac{3x+1}{x^2-x-6} = \frac{c}{x+2} + \frac{d}{x-3}$$

Putting the right hand side under a common denominator gives us $\frac{(c+d)x + (2d-3c)}{x^2-x-6}$.

Good job. All you do is match coefficients: $(c+d)x + (2d-3c)$ must match exactly with $3x+1$.

So our final partial fraction decomposition is $$\frac{3x+1}{x^2-x-6} = \frac{1}{x+2} + \frac{2}{x-3}.$$

This method works pretty well for quadratic denominators, but imagine when the denominator's degree increases! How many equations, variables, and coefficients will we have to match! Below I'll present a more efficient method.

After finding $a, b$ (that was a long time ago!), here is an alternate way of finding $c$ and $d$.

Take the equation $$\frac{3x+1}{x^2-x-6} = \frac{c}{x+2} + \frac{d}{x-3}$$ and multiply both sides by $x+2$. Here is what you get: $$\frac{3x+1}{x-3} = c + \frac{d(x+2)}{x-3}$$

Now plug in $x = -2$ to get $1 = c$ (since the last term cancels out)! It's that easy!

Here's the process again: multiply both sides of the original equation by $x-3$ to get that $\frac{3x+1}{x+2} = \frac{c(x-3)}{x+2} + d$. Now plug in $x = 3$ to get that $2 = d$! Once again, the first term on the right hand side cancels out.

Now let us do a bigger case: for polynomials $p(x)/q(x)$. Note that in this whole process, we'll assume that $\deg p(x) < \deg q(x)$. Or else, divide $p(x)$ into $q(x)$, giving a quotient and a remainder. Then just work with the remainder.

So let's "factor" $q(x)$ as the product of its roots. Let's say that $q(x) = (x-r_1)(x-r_2)\dots(x-r_n)$ ASSUME THAT ALL ROOTS ARE DISTINCT! If not, my above process described actually doesn't work. (Luckily, most things you encounter will have distinct roots)

Then, by our above arguments, we can write

$$\frac{p(x)}{(x-r_1)(x-r_2)\dots(x-r_n)} = \frac{c_1}{x-r_1} + \frac{c_2}{x-r_2} + \dots + \frac{c_n}{x-r_n}$$

Note that the denominator on the left hand side above is in fact $q(x)$. Let's multiply through by $x-r_1$ now.

You get that $$\frac{p(x)}{(x-r_2)(x-r_3) \dots (x-r_n)} \\= c_1 + \frac{c_2(x-r_1)}{x-r_2} + \frac{c_3(x-r_1)}{x-r_3} + \dots + \frac{c_n(x-r_1)}{x-r_n}$$

This is where we have used that every root is distinct. Otherwise, we would be able to simplify other fractions on the right hand side.

Now plug in $x = r_1$. The equation becomes $$\frac{p(r_1)}{(r_1-r_2)(r_1-r_3)\dots (r_1-r_n)} = c_1.$$ So you have your value after plugging in everything!

You can repeat the argument similar for $x-r_i$ to find $c_i$ for all $i$.

In a nicer form, you can actually show that $(r_1-r_2)(r_1-r_3)\dots (r_1-r_n) = q'(r_1)$ (derivative), so in general we can write: $c_i = \frac{p(r_i)}{q'(r_i)}$. But this final formula is not really important: the process and algorithm described are more useful to think about. :)

(Videos) Partial Fraction Decomposition

by PatrickJMT

Here are some videos by patrickJMT showing you how to do partial fraction decomposition.

Partial Fraction Decomposition is the process of taking a simplified fraction and breaking it into smaller fractions.

The first example he looks at is: $$\frac{3x-8}{x^2-4x-5}$$ Before starting the problem there's two things to check:

1. The degree of the numerator always needs to be smaller than the degree of the denominator.
2. Does the denominator factor.

Our denominator can factor so let's go ahead and do that. $$\frac{3x-8}{x^2-4x-5}=\frac{3x-8}{(x-5)(x+1)}$$ Notice that we have two distinct linear factors. What we try to do is break our rational function into two separate fractions. Where each denominator is one of the factors. $$=\frac{A}{x-5}+\frac{B}{x+1}$$ Our job is to figure out the constants $A$ and $B$. There are multiple ways to solve these. Personally, I like the way Patrick goes about it. You take whatever is on the bottom of our starting fraction and multiply it to both sides.

The result, after canceling, is: $$3x-8=A(x+1)+B(x-5)$$ Because our factors are linear, there is a shortcut we can do at this step in order to quickly figure out our $A$ and $B$ values.

The first step is to look at the first term on the right, $A(x+1)$, and ask what value of $x$ can we put in that will cancel that term out? If we pick $x=-1$ then the inside becomes $0$ which removes the term. So now we plug in $x=-1$ into our equation. $$\begin{align} 3(-1)-8&=A(-1+1)+B(-1-5)\\ -11&=-6B\\ \frac{-11}{-6}&=B \end{align}\\ \boxed{B=\frac{11}{6}}$$

Next we do the same thing for the second term on the right. Here, $x=5$ removes the term so let's plug that in. $$\begin{align} 3(5)-8&=A(5+1)+B(5-5)\\ 7&=6A\\ \frac{7}{6}&=A \end{align}\\ \boxed{A=\frac{7}{6}}$$

So now we can write our partial decomposition as $$=\boxed{\frac{\frac76}{x+1}+\frac{\frac{11}6}{x-5}}$$

(Videos) Partial Fraction Decomposition with Nonlinear Factors

by PatrrickJMT

Here's a video by patrickJMT showing you how to take the partial decomposition when there's nonlinear factors.

The first problem he looks at is $$\frac{2x^2+4x-5}{x^2(x+3)}$$ Patrick rewrites the $x$ being in parentheses then squared $$\frac{2x^2+4x-5}{(x)^2(x+3)}$$ The reason he does this is because it's important to think about it being two factors of $x$. Not just a factor of $x^2$. So now we break it down into the following fractions. $$=\frac{A}{x}+\frac{B}{(x)^2}+\frac{C}{x+3}$$

If we had $x^3$ instead of $x^2$ we would have to make a fraction for $x$, $x^2$, and $x^3$.

Now we multiply both sides by the denominator of our original fraction. $$\color{#8925AE}{x^2(x+3)}\frac{2x^2+4x-5}{x^2(x+3)}=\left(\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+3}\right)\color{#8925AE}{x^2(x+3)}$$ Multiplying and canceling factors we get $$2x^2+4x-5=A(x)(x+3)+B(x+3)+C(x^2)$$

Since our factors are not linear we can't use the shortcut from before. Instead we do a process called equating coefficients. Our first step is to multiply everything out. $$2x^2+4x-5=Ax^2+3Ax+Bx+3B+Cx^2$$ Next we group together our like terms. $$2x^2+4x-5=(A+C)x^2+(3A+B)x+3B$$ So now we set the coefficients of each term equal to each other. $$\begin{align} (1)\ \ \ \ \ \ \ \ 2&=A+C\\ (2)\ \ \ \ \ \ \ \ 4&=3A+B\\ (3)\ \ \ -5&=3B \end{align}$$ Now we solve this system to find the values for $A$, $B$, and $C$.

If we look at equation (3) we can easily find the value of $B$. $$-5=3B\\ \frac{-5}3=B\\ \boxed{B=-\frac53}$$ With $B$ we can find $A$ in equation (2). $$4=3A+B\\ 4=3A-\frac53\\ 4+\frac53=3A\\ \frac{17}3=3A\\ \frac{17}9=A\\ \boxed{A=\frac{17}9}$$ Finally, with $A$ we can find $C$ in equation (1). $$2=A+C\\ 2=\frac{17}9+C\\ 2-\frac{17}9=C\\ \frac19=C\\ \boxed{C=\frac19}$$

So the partial decomposition becomes $$=\boxed{\frac{\frac{17}9}{x}+\frac{-\frac53}{x^2}+\frac{\frac19}{x+3}}$$

Partial Fraction Decomposition

Now that we are getting into more complicated math and rational equations, it is important to learn about partial fraction decomposition in order to deal with the large and complex fractions. It may be helpful to review rational expressions, complex numbers, systems of rational equations, and polynomial long division before diving into this lesson.

Partial fraction decomposition allows you to break down large fractions into smaller, more manageable pieces. More importantly, it allows you to solve for variables within the fractions.

The first step is to factor the denominator and replace the numerators with variables. Typically, you will have unknown constants and linear terms in the numerator.

The second step is to set this equal to the original fraction and eliminate the denominators. This will involve a partial fraction expansion, resulting in a system of linear or quadratic equations.

Lastly, the third step is to get rid of the fractions and solve. This is the simplest, most straightforward step.

Looking Further

Many problems in Finite Telescoping Sums and Products begin by using partial fraction decomposition. Breaking each term into simpler fractions makes the cancellation easier to see.