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August 27, 2020

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### Section 5-5 : Partial Fractions

This section doesn’t really have a lot to do with the rest of this chapter, but since the subject needs to be covered and this was a fairly short chapter it seemed like as good a place as any to put it.

So, let’s start with the following. Let’s suppose that we want to add the following two rational expressions.

\begin{align*}\frac{8}{{x + 1}} - \frac{5}{{x - 4}} & = \frac{{8\left( {x - 4} \right)}}{{\left( {x + 1} \right)\left( {x - 4} \right)}} - \frac{{5\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 4} \right)}}\\ & = \frac{{8x - 32 - \left( {5x + 5} \right)}}{{\left( {x + 1} \right)\left( {x - 4} \right)}}\\ & = \frac{{3x - 37}}{{\left( {x + 1} \right)\left( {x - 4} \right)}}\end{align*}

What we want to do in this section is to start with rational expressions and ask what simpler rational expressions did we add and/or subtract to get the original expression. The process of doing this is called partial fractions and the result is often called the partial fraction decomposition.

The process can be a little long and on occasion messy, but it is actually fairly simple. We will start by trying to determine the partial fraction decomposition of,

$\frac{{P\left( x \right)}}{{Q\left( x \right)}}$

where both $$P(x)$$ and $$Q(x)$$ are polynomials and the degree of $$P(x)$$ is smaller than the degree of $$Q(x)$$. Partial fractions can only be done if the degree of the numerator is strictly less than the degree of the denominator. That is important to remember.

So, once we’ve determined that partial fractions can be done we factor the denominator as completely as possible. Then for each factor in the denominator we can use the following table to determine the term(s) we pick up in the partial fraction decomposition.

Factor in
denominator
Term in partial
fraction decomposition
$$ax + b$$ $$\displaystyle \frac{A}{{ax + b}}$$
$${\left( {ax + b} \right)^k}$$ $$\displaystyle \frac{{{A_1}}}{{ax + b}} + \frac{{{A_2}}}{{{{\left( {ax + b} \right)}^2}}} + \cdots + \frac{{{A_k}}}{{{{\left( {ax + b} \right)}^k}}}$$
$$a{x^2} + bx + c$$ $$\displaystyle \frac{{Ax + B}}{{a{x^2} + bx + c}}$$
$${\left( {a{x^2} + bx + c} \right)^k}$$ $$\displaystyle \frac{{{A_1}x + {B_1}}}{{a{x^2} + bx + c}} + \frac{{{A_2}x + {B_2}}}{{{{\left( {a{x^2} + bx + c} \right)}^2}}} + \cdots + \frac{{{A_k}x + {B_k}}}{{{{\left( {a{x^2} + bx + c} \right)}^k}}}$$

Notice that the first and third cases are really special cases of the second and fourth cases respectively if we let $$k = 1$$. Also, it will always be possible to factor any polynomial down into a product of linear factors ($$ax + b$$) and quadratic factors ($$a{x^2} + bx + c$$) some of which may be raised to a power.

There are several methods for determining the coefficients for each term and we will go over each of those as we work the examples. Speaking of which, let’s get started on some examples.

Example 1 Determine the partial fraction decomposition of each of the following.
1. $$\displaystyle \frac{{8x - 42}}{{{x^2} + 3x - 18}}$$
2. $$\displaystyle \frac{{9 - 9x}}{{2{x^2} + 7x - 4}}$$
3. $$\displaystyle \frac{{4{x^2}}}{{\left( {x - 1} \right){{\left( {x - 2} \right)}^2}}}$$
4. $$\displaystyle \frac{{9x + 25}}{{{{\left( {x + 3} \right)}^2}}}$$
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Show Discussion

We’ll go through the first one in great detail to show the complete partial fraction process and then we’ll leave most of the explanation out of the remaining parts.

a $$\displaystyle \frac{{8x - 42}}{{{x^2} + 3x - 18}}$$ Show Solution

The first thing to do is factor the denominator as much as we can.

$\frac{{8x - 42}}{{{x^2} + 3x - 18}} = \frac{{8x - 42}}{{\left( {x + 6} \right)\left( {x - 3} \right)}}$

So, by comparing to the table above it looks like the partial fraction decomposition must look like,

$\frac{{8x - 42}}{{{x^2} + 3x - 18}} = \frac{A}{{x + 6}} + \frac{B}{{x - 3}}$

Note that we’ve got different coefficients for each term since there is no reason to think that they will be the same.

Now, we need to determine the values of $$A$$ and $$B$$. The first step is to actually add the two terms back up. This is usually simpler than it might appear to be. Recall that we first need the least common denominator, but we’ve already got that from the original rational expression. In this case it is,

$LCD = \left( {x + 6} \right)\left( {x - 3} \right)$

Now, just look at each term and compare the denominator to the LCD. Multiply the numerator and denominator by whatever is missing then add. In this case this gives,

$\frac{{8x - 42}}{{{x^2} + 3x - 18}} = \frac{{A\left( {x - 3} \right)}}{{\left( {x + 6} \right)\left( {x - 3} \right)}} + \frac{{B\left( {x + 6} \right)}}{{\left( {x + 6} \right)\left( {x - 3} \right)}} = \frac{{A\left( {x - 3} \right) + B\left( {x + 6} \right)}}{{\left( {x + 6} \right)\left( {x - 3} \right)}}$

We need values of $$A$$ and $$B$$ so that the numerator of the expression on the left is the same as the numerator of the term on the right. Or,

$8x - 42 = A\left( {x - 3} \right) + B\left( {x + 6} \right)$

This needs to be true regardless of the $$x$$ that we plug into this equation. As noted above there are several ways to do this. One way will always work but can be messy and will often require knowledge that we don’t have yet. The other way will not always work, but when it does it will greatly reduce the amount of work required.

In this set of examples, the second (and easier) method will always work so we’ll be using that here. Here we are going to make use of the fact that this equation must be true regardless of the $$x$$ that we plug in.

So, let’s pick an $$x$$, plug it in and see what happens. For no apparent reason let’s try plugging in $$x = 3$$. Doing this gives,

\begin{align*}8\left( 3 \right) - 42 & = A\left( {3 - 3} \right) + B\left( {3 + 6} \right)\\ - 18 & = 9B\\ - 2 & = B\end{align*}

Can you see why we choose this number? By choosing $$x = 3$$ we got the term involving $$A$$ to drop out and we were left with a simple equation that we can solve for $$B$$.

Now, we could also choose $$x = - 6$$ for exactly the same reason. Here is what happens if we use this value of $$x$$.

\begin{align*}8\left( { - 6} \right) - 42 & = A\left( { - 6 - 3} \right) + B\left( { - 6 + 6} \right)\\ - 90 & = - 9A\\ 10 & = A\end{align*}

So, by correctly picking $$x$$ we were able to quickly and easily get the values of $$A$$ and $$B$$. So, all that we need to do at this point is plug them in to finish the problem. Here is the partial fraction decomposition for this part.

$\frac{{8x - 42}}{{{x^2} + 3x - 18}} = \frac{{10}}{{x + 6}} + \frac{{ - 2}}{{x - 3}} = \frac{{10}}{{x + 6}} - \frac{2}{{x - 3}}$

Notice that we moved the minus sign on the second term down to make the addition a subtraction. We will always do that.

b $$\displaystyle \frac{{9 - 9x}}{{2{x^2} + 7x - 4}}$$ Show Solution

Okay, in this case we won’t put quite as much detail into the problem. We’ll first factor the denominator and then get the form of the partial fraction decomposition.

$\frac{{9 - 9x}}{{2{x^2} + 7x - 4}} = \frac{{9 - 9x}}{{\left( {2x - 1} \right)\left( {x + 4} \right)}} = \frac{A}{{2x - 1}} + \frac{B}{{x + 4}}$

In this case the LCD is $$\left( {2x - 1} \right)\left( {x + 4} \right)$$ and so adding the two terms back up give,

$\frac{{9 - 9x}}{{2{x^2} + 7x - 4}} = \frac{{A\left( {x + 4} \right) + B\left( {2x - 1} \right)}}{{\left( {2x - 1} \right)\left( {x + 4} \right)}}$

Next, we need to set the two numerators equal.

$9 - 9x = A\left( {x + 4} \right) + B\left( {2x - 1} \right)$

Now all that we need to do is correctly pick values of $$x$$ that will make one of the terms zero and solve for the constants. Note that in this case we will need to make one of them a fraction. This is fairly common so don’t get excited about it. Here is this work.

\begin{align*}x = - 4: & \hspace{0.25in}45 = - 9B & \Rightarrow \hspace{0.25in}B & = - 5\\ x = \frac{1}{2}:& \hspace{0.25in}\frac{9}{2} = A\left( {\frac{9}{2}} \right) & \Rightarrow \hspace{0.25in}A & = 1\end{align*}

The partial fraction decomposition for this expression is,

$\frac{{9 - 9x}}{{2{x^2} + 7x - 4}} = \frac{1}{{2x - 1}} - \frac{5}{{x + 4}}$

c $$\displaystyle \frac{{4{x^2}}}{{\left( {x - 1} \right){{\left( {x - 2} \right)}^2}}}$$ Show Solution

In this case the denominator has already been factored for us. Notice as well that we’ve now got a linear factor to a power. So, recall from our table that this means we will get 2 terms in the partial fraction decomposition from this factor. Here is the form of the partial fraction decomposition for this expression.

$\frac{{4{x^2}}}{{\left( {x - 1} \right){{\left( {x - 2} \right)}^2}}} = \frac{A}{{x - 1}} + \frac{B}{{x - 2}} + \frac{C}{{{{\left( {x - 2} \right)}^2}}}$

Now, remember that the LCD is just the denominator of the original expression so in this case we’ve got $$\left( {x - 1} \right){\left( {x - 2} \right)^2}$$. Adding the three terms back up gives us,

$\frac{{4{x^2}}}{{\left( {x - 1} \right){{\left( {x - 2} \right)}^2}}} = \frac{{A{{\left( {x - 2} \right)}^2} + B\left( {x - 1} \right)\left( {x - 2} \right) + C\left( {x - 1} \right)}}{{\left( {x - 1} \right){{\left( {x - 2} \right)}^2}}}$

Remember that we just need to add in the factors that are missing to each term.

Now set the numerators equal.

$4{x^2} = A{\left( {x - 2} \right)^2} + B\left( {x - 1} \right)\left( {x - 2} \right) + C\left( {x - 1} \right)$

In this case we’ve got a slightly different situation from the previous two parts. Let’s start by picking a couple of values of $$x$$ and seeing what we get since there are two that should jump right out at us as being particularly useful.

\begin{align*}x = 1: & \hspace{0.25in}4 = A{\left( { - 1} \right)^2} & \Rightarrow \hspace{0.25in}A & = 4\\ x = 2:&\hspace{0.25in}16 = C\left( 1 \right) & \Rightarrow \hspace{0.25in}C & = 16\end{align*}

So, we can get $$A$$ and $$C$$ in the same manner that we’ve been using to this point. However, there is no value of $$x$$ that will allow us to eliminate the first and third term leaving only the middle term that we can use to solve for $$B$$. While this may appear to be a problem it actually isn’t. At this point we know two of the three constants. So all we need to do is chose any other value of $$x$$ that would be easy to work with ($$x = 0$$ seems particularly useful here), plug that in along with the values of $$A$$ and $$C$$ and we’ll get a simple equation that we can solve for $$B$$.

Here is that work.

\begin{align*}4{\left( 0 \right)^2} & = \left( 4 \right){\left( { - 2} \right)^2} + B\left( { - 1} \right)\left( { - 2} \right) + 16\left( { - 1} \right)\\ 0 & = 16 + 2B - 16\\ 0 & = 2B\\ 0 & = B\end{align*}

In this case we got $$B = 0$$ this will happen on occasion, but do not expect it to happen in all cases. Here is the partial fraction decomposition for this part.

$\frac{{4{x^2}}}{{\left( {x - 1} \right){{\left( {x - 2} \right)}^2}}} = \frac{4}{{x - 1}} + \frac{{16}}{{{{\left( {x - 2} \right)}^2}}}$

d $$\displaystyle \frac{{9x + 25}}{{{{\left( {x + 3} \right)}^2}}}$$ Show Solution

Again, the denominator has already been factored for us. In this case the form of the partial fraction decomposition is,

$\frac{{9x + 25}}{{{{\left( {x + 3} \right)}^2}}} = \frac{A}{{x + 3}} + \frac{B}{{{{\left( {x + 3} \right)}^2}}}$

Adding the two terms together gives,

$\frac{{9x + 25}}{{{{\left( {x + 3} \right)}^2}}} = \frac{{A\left( {x + 3} \right) + B}}{{{{\left( {x + 3} \right)}^2}}}$

Notice that in this case the second term already had the LCD under it and so we didn’t need to add anything in that time.

Setting the numerators equal gives,

$9x + 25 = A\left( {x + 3} \right) + B$

Now, again, we can get $$B$$ for free by picking $$x = - 3$$.

\begin{align*}9\left( { - 3} \right) + 25 & = A\left( { - 3 + 3} \right) + B\\ - 2 & = B\end{align*}

To find $$A$$ we will do the same thing that we did in the previous part. We’ll use $$x = 0$$ and the fact that we know what $$B$$ is.

\begin{align*}25 & = A\left( 3 \right) - 2\\ 27 & = 3A\\ 9 & = A\end{align*}

In this case, notice that the constant in the numerator of the first isn’t zero as it was in the previous part. Here is the partial fraction decomposition for this part.

$\frac{{9x + 25}}{{{{\left( {x + 3} \right)}^2}}} = \frac{9}{{x + 3}} - \frac{2}{{{{\left( {x + 3} \right)}^2}}}$

Now, we need to do a set of examples with quadratic factors. Note however, that this is where the work often gets fairly messy and in fact we haven’t covered the material yet that will allow us to work many of these problems. We can work some simple examples however, so let’s do that.

Example 2 Determine the partial fraction decomposition of each of the following.
1. $$\displaystyle \frac{{8{x^2} - 12}}{{x\left( {{x^2} + 2x - 6} \right)}}$$
2. $$\displaystyle \frac{{3{x^3} + 7x - 4}}{{{{\left( {{x^2} + 2} \right)}^2}}}$$
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a $$\displaystyle \frac{{8{x^2} - 12}}{{x\left( {{x^2} + 2x - 6} \right)}}$$ Show Solution

In this case the $$x$$ that sits in the front is a linear term since we can write it as,

$x = x + 0$

and so the form of the partial fraction decomposition is,

$\frac{{8{x^2} - 12}}{{x\left( {{x^2} + 2x - 6} \right)}} = \frac{A}{x} + \frac{{Bx + C}}{{{x^2} + 2x - 6}}$

Now we’ll use the fact that the LCD is $$x\left( {{x^2} + 2x - 6} \right)$$ and add the two terms together,

$\frac{{8{x^2} - 12}}{{x\left( {{x^2} + 2x - 6} \right)}} = \frac{{A\left( {{x^2} + 2x - 6} \right) + x\left( {Bx + C} \right)}}{{x\left( {{x^2} + 2x - 6} \right)}}$

Next, set the numerators equal.

$8{x^2} - 12 = A\left( {{x^2} + 2x - 6} \right) + x\left( {Bx + C} \right)$

This is where the process changes from the previous set of examples. We could choose $$x = 0$$ to get the value of $$A$$, but that’s the only constant that we could get using this method and so it just won’t work all that well here.

What we need to do here is multiply the right side out and then collect all the like terms as follows,

\begin{align*}8{x^2} - 12 &= A{x^2} + 2Ax - 6A + B{x^2} + Cx\\ 8{x^2} - 12 & = \left( {A + B} \right){x^2} + \left( {2A + C} \right)x - 6A\end{align*}

Now, we need to choose $$A$$, $$B$$, and $$C$$ so that these two are equal. That means that the coefficient of the x2 term on the right side will have to be 8 since that is the coefficient of the x2 term on the left side. Likewise, the coefficient of the $$x$$ term on the right side must be zero since there isn’t an $$x$$ term on the left side. Finally the constant term on the right side must be -12 since that is the constant on the left side.

We generally call this setting coefficients equal and we’ll write down the following equations.

\begin{align*}A + B & = 8\\ 2A + C & = 0\\ - 6A & = - 12\end{align*}

Now, we haven’t talked about how to solve systems of equations yet, but this is one that we can do without that knowledge. We can solve the third equation directly for $$A$$ to get that $$A = 2$$. We can then plug this into the first two equations to get,

\begin{align*}2 + B & = 8 & \Rightarrow \hspace{0.25in}B & = 6\\ 2\left( 2 \right) + C & = 0 & \Rightarrow \hspace{0.25in} C & = - 4\end{align*}

So, the partial fraction decomposition for this expression is,

$\frac{{8{x^2} - 12}}{{x\left( {{x^2} + 2x - 6} \right)}} = \frac{2}{x} + \frac{{6x - 4}}{{{x^2} + 2x - 6}}$

b $$\displaystyle \frac{{3{x^3} + 7x - 4}}{{{{\left( {{x^2} + 2} \right)}^2}}}$$ Show Solution

Here is the form of the partial fraction decomposition for this part.

$\frac{{3{x^3} + 7x - 4}}{{{{\left( {{x^2} + 2} \right)}^2}}} = \frac{{Ax + B}}{{{x^2} + 2}} + \frac{{Cx + D}}{{{{\left( {{x^2} + 2} \right)}^2}}}$

Adding the two terms up gives,

$\frac{{3{x^3} + 7x - 4}}{{{{\left( {{x^2} + 2} \right)}^2}}} = \frac{{\left( {Ax + B} \right)\left( {{x^2} + 2} \right) + Cx + D}}{{{{\left( {{x^2} + 2} \right)}^2}}}$

Now, set the numerators equal and we might as well go ahead and multiply the right side out and collect up like terms while we’re at it.

\begin{align*}3{x^3} + 7x - 4 & = \left( {Ax + B} \right)\left( {{x^2} + 2} \right) + Cx + D\\ 3{x^3} + 7x - 4 & = A{x^3} + 2Ax + B{x^2} + 2B + Cx + D\\ 3{x^3} + 7x - 4 & = A{x^3} + B{x^2} + \left( {2A + C} \right)x + 2B + D\end{align*}

Setting coefficients equal gives,

\begin{align*}A & = 3\\ B & = 0\\ 2A + C & = 7\\ 2B + D & = - 4\end{align*}

In this case we got $$A$$ and $$B$$ for free and don’t get excited about the fact that $$B = 0$$. This is not a problem and in fact when this happens the remaining work is often a little easier. So, plugging the known values of $$A$$ and $$B$$ into the remaining two equations gives,

\begin{align*}2\left( 3 \right) + C & = 7 &\Rightarrow \hspace{0.25in} C & = 1\\ 2\left( 0 \right) + D & = - 4 &\Rightarrow \hspace{0.25in} D & = - 4\end{align*}

The partial fraction decomposition is then,

$\frac{{3{x^3} + 7x - 4}}{{{{\left( {{x^2} + 2} \right)}^2}}} = \frac{{3x}}{{{x^2} + 2}} + \frac{{x - 4}}{{{{\left( {{x^2} + 2} \right)}^2}}}$
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