For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Partial Fractions
 Long ago, you learned how to combine fractions: you put them over a common denominator, then combine them. But what if we wanted to do the reverse? What if we had a fraction that we wanted to break up into multiple, smaller fractions? We call these smaller fractions partial fractions and the process partial fraction decomposition.
 To understand this lesson, you'll need some familiarity with solving systems of linear equations. Previous experience from past Algebra classes will probably be enough, but if you want a refresher, check out the lesson Systems of Linear Equations. Knowing how to factor polynomials and understanding them in general will also be necessary, along with the ability to do polynomial division.
 If we have a polynomial fraction of the from [N(x)/D(x)], there are two possibilities in regards to the degrees of the top and bottom polynomials:
 Proper: degree of N(x) < degree of D(x);
 Improper: degree of N(x) ≥ degree of D(x).
 Once the polynomial fraction is proper, the next step is to factor the denominator. After it's broken down into its smallest factors, we're ready to decompose. Notice that there are two types that these smallest possible factors can come in:
 Linear factors raised to a power: (ax+b)^{m};
 Irreducible quadratic factors raised to a power: (ax^{2}+bx+c)^{m}. [Remember, `irreducible' means it can't be broken up further. That means quadratics like x^{2}+1 or 5x^{2}−3x+20.]
 The partial fraction decomposition must include the following for each linear factor (ax+b)^{m}:
where A_{1}, …, A_{m} are all constant real numbers.A_{1} ax+b
+ A_{2} (ax+b)^{2}
+ …+ A_{m} (ax+b)^{m}
,  The partial fraction decomposition must include the following for each irreducible quadratic (ax^{2}+bx+c)^{m}:
where A_{1}, …, A_{m} and B_{1}, …, B_{m} are all constants.A_{1}x+B_{1} ax^{2}+bx+c
+ A_{2}x+B_{2} (ax^{2}+bx+c)^{2}
+ …+ A_{m}x+B_{m} (ax^{2}+bx+c)^{m}
,  If we have multiple of a given type or mixed types, we just decompose each of them based on their rules and put them all together.
 Finally, we need to figure out what goes on the numerators. To do this, have the original fraction on one side of an equation and the partial fraction decomposition on the other. Then multiply both sides of the equation by the denominator of the original fraction. This gives an equation that we can transform into a system of linear equations and solve to find the values for each of the constants.
Note:Before starting, it should be noted that this is a rather difficult concept to explain just with writing. It helps a lot to see it in action, so it's strongly recommended that you watch the video if you find any of this confusing.
Partial Fractions
 A rational function is a fraction made out of polynomials: the numerator and the denominator are both polynomials. This means we can talk about the degree (largest exponent on a variable) of each of the two polynomials.
 A fraction is proper when the degree of the numerator is less than the degree of the denominator.
 A fraction is improper when the degree of the numerator is equal to or greater than the degree of the denominator.
 These two definitions matter for partial fraction decomposition because only proper fractions can be decomposed. If a fraction is in improper form, it must first me put into proper form through polynomial division. Only then can it be decomposed.

 To decompose a fraction, the fraction must be proper. For this problem, we see that the numerator has a smaller degree than the denominator, so we're fine to continue. Next, we need to factor the denominator completely. This has already been done for us.
 Once the denominator is factored, we can create an equation where the lefthand side is the original fraction and the righthand side uses the denominator's factors and currently unknown constants. [See the lesson if you're not sure how to do this; it is fully explained in the lesson and much easier to understand visually.]
3x−2 (x+1)(x−3)= A x+1+ B x−3

 To decompose a fraction, the fraction must be proper. For this problem, we see that the numerator has a smaller degree than the denominator, so we're fine to continue. Next, we need to factor the denominator completely. This has already been done for us.
 Once the denominator is factored, we can create an equation where the lefthand side is the original fraction and the righthand side uses the denominator's factors and currently unknown constants. [See the lesson if you're not sure how to do this; it is fully explained in the lesson and much easier to understand visually.]
2x+6 x(x+2)= A x+ B x+2  To solve for the constants on top of the fractions, multiply both sides by the factors in our starting fraction's denominator:
x(x+2) · ⎡
⎣2x+6 x(x+2)= A x+ B x+2⎤
⎦2x+6 = A(x+2) + B(x) 2x+6 = Ax + Bx + 2A  From here, we can solve based on each type. The only things that can contribute to the number of x's are terms with an x on them, and the only thing that can contribute to the constant number are terms with no variables on them. Thus, we can split the above into two equations:
2x = Ax + Bx 6 = 2A  We can easily solve the rightside equation and get A = 3. Once we know that, we can plug it into the other equation to find B:
so B=−1.2x = 3x + Bx,  Now plug our values of A=3 and B=−1 into the decomposition we set up near the beginning to find how the fraction breaks down:
2x+6 x(x+2)= A x+ B x+2= 3 x+ −1 x+2

 To decompose a fraction, the fraction must be proper. For this problem, we see that the numerator has a smaller degree than the denominator, so we're fine to continue. Next, we need to factor the denominator completely. This has already been done for us.
 Once the denominator is factored, we can create an equation where the lefthand side is the original fraction and the righthand side uses the denominator's factors and currently unknown constants. [See the lesson if you're not sure how to do this; it is fully explained in the lesson and much easier to understand visually.]
4x−3 (x−6)(x+1)= A x−6+ B x+1  To solve for the constants on top of the fractions, multiply both sides by the factors in our starting fraction's denominator:
(x−6)(x+1) · ⎡
⎣4x−3 (x−6)(x+1)= A x−6+ B x+1⎤
⎦4x−3 = A(x+1) + B(x−6) 4x−3 = Ax + Bx + A−6B  From here, we can solve based on each type. The only things that can contribute to the number of x's are terms with an x on them, and the only thing that can contribute to the constant number are terms with no variables on them. Thus, we can split the above into two equations:
4x = Ax + Bx −3 = A−6B  While we can't solve for any numbers yet, we can put A in terms of B: A = 6B−3. We can then plug that into the other equation.
which we can then solve to find B=1.4x = (6B−3)x + Bx,  Now that we know B=1, we can easily find A:
A = 6 (1)−3 ⇒ A = 3  Now plug our values of A=3 and B=1 into the decomposition we set up near the beginning to find how the fraction breaks down:
4x−3 (x−6)(x+1)= A x−6+ B x+1= 3 x−6+ 1 x+1

 To decompose a fraction, the fraction must be proper. For this problem, we see that the numerator has a smaller degree than the denominator, so we're fine to continue. Next, we need to factor the denominator completely. This hasn't been done, so we factor x^{2}+2x−35, allowing us to rewrite the fraction as
11−7x x^{2}+2x−35= 11−7x (x+7)(x−5)  Once the denominator is factored, we can create an equation where the lefthand side is the original fraction and the righthand side uses the denominator's factors and currently unknown constants. [See the lesson if you're not sure how to do this; it is fully explained in the lesson and much easier to understand visually.]
11−7x (x+7)(x−5)= A x+7+ B x−5  To solve for the constants on top of the fractions, multiply both sides by the factors in our starting fraction's denominator:
(x+7)(x−5) · ⎡
⎣11−7x (x+7)(x−5)= A x+7+ B x−5⎤
⎦11−7x = A(x−5) + B(x+7) 11−7x = Ax + Bx−5A+7B  From here, we can solve based on each type. The only things that can contribute to the number of x's are terms with an x on them, and the only thing that can contribute to the constant number are terms with no variables on them. Thus, we can split the above into two equations:
−7x = Ax + Bx 11 = −5A+7B −7 = A+B 11 = −5A + 7B  While we can't solve for any numbers yet, we can put B in terms of A: B = −7−A. We can then plug that into the other equation.
which we can then solve to find A=−5.11 = −5A + 7(−7−A),  Now that we know A=−5, we can easily find B:
B = −7−(−5) ⇒ B = −2  Now plug our values of A=−5 and B=−2 into the decomposition we set up near the beginning to find how the fraction breaks down:
11−7x (x+7)(x−5)= A x+7+ B x−5= −5 x+7+ −2 x−5

 To decompose a fraction, the fraction must be proper. For this problem, we see that the numerator has a smaller degree than the denominator, so we're fine to continue. Next, we need to factor the denominator completely. This has already been done for us. [Notice that (x^{2}+1) cannot be factored further because it is irreducible, so the denominator has been factored completely.]
 Once the denominator is factored, we can create an equation where the lefthand side is the original fraction and the righthand side uses the denominator's factors and currently unknown constants. [See the lesson if you're not sure how to do this; it is fully explained in the lesson and much easier to understand visually.] Remember, if a factor is an irreducible quadratic (like x^{2}+1), instead of a single constant on the top, it gets Ax+B. Also remember, if there are multiple of a single factor (like (x+7)^{2}), a fraction appears for each multiple, and at each "exponent level".
x^{2}−7 (x^{2}+1)(x+7)^{2}= Ax+B x^{2}+1+ C x+7+ D (x+7)^{2}

 To decompose a fraction, the fraction must be proper. It would be difficult to expand the denominator, but we can see that its degree must be greater than the numerator's if we look at how much each factor would contribute to the overall degree. Next, we need to factor the denominator completely. It looks like this has already been done for us, but we should check to make sure that (3x^{2}−5x+4) is actually irreducible. We can do so with the discriminant:
thus it is indeed irreducible, so the denominator is completely factored.b^{2}−4ac = (−5)^{2} − 4(3)(4) = −23 < 0,  Once the denominator is factored, we can create an equation where the lefthand side is the original fraction and the righthand side uses the denominator's factors and currently unknown constants. [See the lesson if you're not sure how to do this; it is fully explained in the lesson and much easier to understand visually.] Remember, if a factor is an irreducible quadratic (like 3x^{2}−5x+4), instead of a single constant on the top, it gets Ax+B. Also remember, if there are multiple of a single factor, a fraction appears for each multiple, and at each "exponent level".
4x^{6}+8x^{3}−27x^{2}+10 (x+2)(x−1)^{3}(3x^{2}−5x+4)^{2}= A x+2+ B x−1+ C (x−1)^{2}+ D (x−1)^{3}+ Ex+F 3x^{2}−5x+4+ Gx+H (3x^{2}−5x+4)^{2}

 To decompose a fraction, the fraction must be proper. For this problem, we see that the numerator has a smaller degree than the denominator, so we're fine to continue. Next, we need to factor the denominator completely. This has already been done for us. [Notice that (x^{2}+3) is an irreducible quadratic.]
 Now we can follow the rules of decomposition to set up the equation we'll work with:
−7x−21 (x−5)(x^{2}+3)= A x−5+ Bx+C x^{2}+3  To solve for the constants on top of the fractions, multiply both sides by the factors in our starting fraction's denominator:
(x−5)(x^{2}+3) · ⎡
⎣−7x−21 (x−5)(x^{2}+3)= A x−5+ Bx+C x^{2}+3⎤
⎦−7x−21 = A(x^{2}+3) + (Bx+C)(x−5) −7x−21 = Ax^{2} + Bx^{2} −5Bx+Cx+ 3A−5C  From here, we can solve based on each type. We effectively have three different types: x^{2}, x, and constant. Thus, we can break the above equation into three separate equations:
0 = A + B −7 = −5B + C −21 = 3A −5C  We will solve through substitution. From the above, we can quickly see that B = −A. Plugging that in, we have
We can then plug that in to the last remaining equation to get−7 = −5 (−A) + C ⇒ C = −5A − 7. −21 = 3A −5(−5A−7) ⇒ −21 = 28A +35 ⇒ A = −2  Now that we know A=−2, we can work out the others:
B = −(−2) ⇒ B = 2 C = −5(−2)−7 ⇒ C = 3  Now plug our values of A=−2, B=2, and C=3 into the decomposition we set up near the beginning to find how the fraction breaks down:
−7x−21 (x−5)(x^{2}+3)= A x−5+ Bx+C x^{2}+3= −2 x−5+ 2x+3 x^{2}+3

 To decompose a fraction, the fraction must be proper. For this problem, we see that the numerator has a smaller degree than the denominator, so we're fine to continue. Next, we need to factor the denominator completely. This hasn't been done, so we factor x^{4}+10x^{2}+25, allowing us to rewrite the fraction as
[Notice that (x^{2}+5) is irreducible, so we're done factoring.]3x^{3}−4x^{2}+16x−27 x^{4}+10x^{2}+25= 3x^{3}−4x^{2}+16x−27 (x^{2}+5)^{2}  Now we can follow the rules of decomposition to set up the equation we'll work with:
3x^{3}−4x^{2}+16x−27 (x^{2}+5)^{2}= Ax+B x^{2}+5+ Cx+D (x^{2}+5)^{2}  To solve for the constants on top of the fractions, multiply both sides by the factors in our starting fraction's denominator:
(x^{2}+5)^{2} · ⎡
⎣3x^{3}−4x^{2}+16x−27 (x^{2}+5)^{2}= Ax+B x^{2}+5+ Cx+D (x^{2}+5)^{2}⎤
⎦3x^{3}−4x^{2}+16x−27 = (Ax+B)(x^{2}+5) + (Cx+D) 3x^{3}−4x^{2}+16x−27 = Ax^{3} + Bx^{2} + 5Ax + Cx+5B+ D  From here, we can solve based on each type. We effectively have four different types: x^{3}, x^{2}, x, and constant. Thus, we can break the above equation into four separate equations:
3 = A −4 = B 16 = 5A + C −27 = 5B + D  We will finish solving through substitution. Plug in, using what we found above:
16 = 5(3) + C ⇒ C = 1. −27 = 5(−4)+D ⇒ D=−7  Now plug our values of A=3, B=−4, C=1, and D=−7 into the decomposition we set up near the beginning to find how the fraction breaks down:
3x^{3}−4x^{2}+16x−27 (x^{2}+5)^{2}= Ax+B x^{2}+5+ Cx+D (x^{2}+5)^{2}= 3x−4 x^{2}+5+ x−7 (x^{2}+5)^{2}

 To decompose a fraction, the fraction must be proper. However, for this problem, the fraction is improper because the numerator's degree (3) is greater than the denominator's degree (2). We can break the fraction into a nonfraction and proper fraction by using polynomial long division (if you're unfamiliar with polynomial long division, check out the lesson on it):
x^{2} −3x −10 ⎞
⎠3x^{3} −7x^{2} −32x +9
Thus, by long division, we have shown that3x +2 R: (4x+29) x^{2} −3x −10 ⎞
⎠3x^{3} −7x^{2} −32x +9 3x^{3} −9x^{2} −30x 2x^{2} −2x + 9 2x^{2} −6x −20 4x +29 3x^{3}−7x^{2}−32x+9 x^{2}−3x−10= 3x + 2 + 4x+29 x^{2}−3x−10 At this point, we can now focus on decomposing the proper fraction from what we just figured out. We need to make sure we don't forget about the 3x+2 in front of the fraction we're decomposing, so make a note to put it back in at the end in front of whatever decomposition we figure out.
 Now that we have a proper fraction, we need to factor the denominator completely. This hasn't been done yet, so we factor x^{2}−3x−10, allowing us to rewrite the fraction we're working with as
4x+29 x^{2}−3x−10= 4x+29 (x+2)(x−5)  Once the denominator is factored, we can create an equation where the lefthand side is the fraction and the righthand side uses the denominator's factors and currently unknown constants. [See the lesson if you're not sure how to do this; it is fully explained in the lesson and much easier to understand visually.]
4x+29 (x+2)(x−5)= A x+2+ B x−5  To solve for the constants on top of the fractions, multiply both sides by the factors in our fraction's denominator:
(x+2)(x−5) · ⎡
⎣4x+29 (x+2)(x−5)= A x+2+ B x−5⎤
⎦4x+29 = A(x−5) + B(x+2) 4x+29 = Ax + Bx−5A+2B  From here, we can split the above into two equations (based on whether or not there's a variable attached), and work to solve from there:
4 = A + B 29 = −5A+2B  Use substitution to solve. We find that B = 4−A, so plug that in:
which we can then solve to find A=−3.29 = −5A + 2(4−A)  Now that we know A=−3, we can easily find B:
B = 4−(−3) ⇒ B = 7  Now plug our values of A=−3 and B=7 into the decomposition we set up to find how the fraction breaks down:
4x+29 (x+2)(x−5)= A x+2+ B x−5= −3 x+2+ 7 x−5  Finally, don't forget that we started with a different fraction that we used long division on. We started with
and now, through the decomposition we just did, we've shown that3x^{3}−7x^{2}−32x+9 x^{2}−3x−10= 3x + 2 + 4x+29 x^{2}−3x−10, 3x^{3}−7x^{2}−32x+9 x^{2}−3x−10= 3x+2 + −3 x+2+ 7 x−5
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Partial Fractions
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro 0:00
 Introduction: Idea 0:04
 Introduction: Prerequisites and Uses 1:57
 Proper vs. Improper Polynomial Fractions 3:11
 Possible Things in the Denominator 4:38
 Linear Factors 6:16
 Example of Linear Factors
 Multiple Linear Factors
 Irreducible Quadratic Factors 8:25
 Example of Quadratic Factors
 Multiple Quadratic Factors
 Mixing Factor Types 10:28
 Figuring Out the Numerator 11:10
 How to Solve for the Constants
 Quick Example
 Example 1 14:29
 Example 2 18:35
 Example 3 20:33
 Example 4 28:51
Precalculus with Limits Online Course
Transcription: Partial Fractions
Hiwelcome back to Educator.com.0000
Today, we are going to talk about partial fractions.0002
Consider if we had two quotients of polynomials that we wanted to add together, like, say, 7/(x  5) and 3/(x + 2).0005
We could put them over a common denominator, then combine them; we have x  5 here and x + 2 here,0014
so we can multiply the one on the left by x + 2, so we get (x  5)(x + 2) on the bottom.0019
Its top will get also multiplied by that (x + 2).0026
For the other one, we will have (x  5) multiplied on it, so it has (x  5) multiplied on the top,0030
so we have (x  5)(x + 2)a common denominatoron the bottom now.0035
We multiply this out; we get 7x + 14 and 3x  15; we combine those, and we get 10x  1 on top.0039
divided by (x  5)(x + 2), expanded to x^{2}  3x  10.0049
So, it is not a bad idea; we can get from here to here by putting it over a common denominator, and then just adding things out and simplifying.0054
But if we wanted to do the reverse processwhat if we started with a fraction involving large polynomials,0064
and we wanted to break it into smaller fractions made of the polynomials' factors?0070
If we wanted to do this process in reverse, if we started at (10x  1)/(x^{2}  3x  10),0074
and we somehow wanted to be able to get that into 7/(x  5) and 3/(x + 2), what process could we go through?0080
We call the smaller fractions on the right partial fractions, because they are parts of that larger fraction.0088
So, each of these here (7/(x  5) and 3/(x + 2))they are each called a partial fraction.0095
And the process to break it up is called partial fraction decomposition.0101
So, whatever this method is that we haven't explored yet, to get from that larger polynomial quotient0105
into these smaller partial fractions, is called partial fraction decomposition.0111
So, how do we do it? To understand this lessonto understand how we do ityou will need some familiarity with solving systems of linear equations.0116
Previous experience from past algebra classes will probably be enough to get through this.0124
But if you want a refresher, if you are a little confused by how this stuff is working later on,0128
check out the lesson Systems of Linear Equationsthat will help explain this stuff if it confuses you right now.0132
You also need to know how to factor polynomials, and you will need to understand them in general,0138
along with the ability to do polynomial division; you will need all of that for some of the stuff in here, as well.0141
Now, sadly, we won't be able to see the application of partial fraction decomposition in this course.0147
However, it is quite useful in calculus, where it will allow us to solve otherwise impossible problems.0152
Partial fraction decomposition is this really useful thing that lets us break up these complicated things0157
that we couldn't solve, and turn them into a form that is actually pretty easy to solve.0162
It is really handy in calculus.0166
We won't be able to see its use in this course; we won't see it any time soon.0167
But it is helpful to practice it now, just like we practiced stuff about factoring complicated polynomials in algebra,0171
before we really had a great understanding of what it was all about.0176
We are practicing something so that we can use it later on.0179
And also, it just is a great chance to flex our brain and get some mental "muscles" that are kind of difficult ideas, but really require some analytic thought.0182
All right, let's get to actually figuring out how to do this.0191
If we have a polynomial fraction in the form numerator polynomial divided by denominator polynomial,0194
a normal rational function format, there are two possibilities in regards to the degrees of the top and bottom polynomials.0199
We call them proper (proper is when the degree of the numerator polynomial is less than the degree of the denominator polynomial0206
the degree of n is less than the degree of the denominator) and improper (when the degree of our numerator0214
is greater than or equal to the degree of the denominator).0222
To decompose the fraction, it must be proper; we have to be in this proper format.0226
We have to make sure that the degree of our numerator is less than the degree of our denominator, if we want to do partial fraction decomposition.0231
So, what do we do if the fraction is improper?0239
We have to turn it into a proper thing; so if the fraction is improper, and we want to decompose it,0241
we have to make it proper through polynomial division.0245
We can make that numerator smaller by being able to break it off and divide out the part that isn't smaller and the part that is smaller.0249
Remember, when you divide your polynomial division, the remainder from the division0256
goes back onto our original denominator, which we will then decompose.0261
We will be able to decompose the part that is the remainder.0265
The other part that comes out cleanlywell, that is just going to be there.0267
That is just going to be a polynomial that is then going to be added to whatever ends up coming out of our decomposition.0271
We will see an example of this in Example 3.0276
Once we have a proper polynomial fraction to decompose into partial fractions0280
(remember, proper: our numerator has to be smaller than our denominator before we can enact this process),0284
once we have done that, we can then factor the denominator.0289
Our next step is to factor the denominator.0293
After the denominator is broken into its smallest possible factors, we are ready to decompose.0295
Now, there are two types that the smallest possible factors can come in.0300
They can come in linear factors, which is forms ax + b, and they will be raised to some power, because we might have a multiplicity of these factors.0302
There might be multiple of a given factor, so if we have multiple ax + b's, we will have m of them, (ax + b)^{m}.0309
So, we will have linear factors in this; or we could have irreducible quadratic factors,0316
that is, things ax^{2} + bx + c that can't be broken up further into linear factors.0320
There is no way to break them up further in the reals.0325
So, ax^{2} + bx + c...and once again, it will be raised to some power;0328
there will be m of them, so it is raised to the m, because there are m of them multiplying together.0331
Remember: "irreducible" means it can't be broken up further in the reals.0336
That means quadratics like x^{2} + 1 or 5x^{2}  3x + 20, because they have no roots.0339
x^{2} + 1 has no roots; x^{2} + 1 = 0...there are no solutions, in the reals, at least.0346
If we allow for the complex, there are; but we are not working with the complex.0355
There are no solutions in the reals; so, since there are no solutions in the reals, it can't be factored any further.0358
x^{2} + 1 is irreducible, therefore.0364
The next step of decomposition will behave differently, depending on which flavor of factor we have0366
whether we are at a linear factor or we are at an irreducible factor; and we will look at the two, one after another.0371
Linear factors: the partial fraction decomposition must include the following for each linear factor that is in this form (ax + b)^{m}.0377
So, it is going to have this in its decomposition: A_{1}/(ax + b) + A_{2}/(ax + b)^{2} +...0385
notice that we are doing this where it is going to keep stepping up with this exponent on the bottom,0394
as we keep putting these things in, until eventually we get up to our m^{th} step,0398
because we had m of them to begin with, so we have to have m of them in the end.0402
And each one of these A's on top (A_{1} up until A_{m}) are all just constant real numbers.0406
We are using A with a subscript, this A_{1}, A_{2}, A_{3} business,0411
because we just need a way of being able to say m of them, and we are not quite sure that we are going to go only out to m.0415
Anyway, the point is that they are all going to be constant real numbers.0420
For example, if we had (x + 7)^{3}, then we have some stuff up top.0423
We don't really care what the stuff is right now, because we just want to see how it will break out.0428
We will deal with the stuff later.0431
If we have stuff over (x + 7)^{3}, then it is going to break into 3 of these, because of this "cubed."0433
We will have A/(x + 7) + B (we can switch into just using letters in general, because we know0439
that each one of these capital letters just means some constant real number; we will figure them out later0444
that will come up laterdon't worry)...A/(x + 7) + B/(x + 7)^{2} + C/(x + 7)^{3}.0449
Notice: we started cubed, and so we have three of these; we step up each time with the exponent,0457
until we have gotten to the number of our multiplicity that we originally started with0462
whatever the exponent was originally on the factor.0465
If we have multiple linear factors, we do it for each one of these; so if we have (2x + 3)^{2}, x^{1}0469
(since there is nothing there), (x  5)^{1} (since there is nothing there)...we have A/(2x + 3) + B/(2x + 3)^{2}0475
(because remember: it was squared to begin with, so it has to have 2 of itself) + C/x.0483
And there is only to the 1, so there is only going to be one of it; plus D/(x  5).0490
And it is only 1, because there is only one of them to begin with.0495
And each one of these A, B, C, D...they are each just constant numbers.0498
We are going to figure them out later on; don't worry.0503
Irreducible quadratic factors: the method is very similar for irreducible quadratic factors.0507
The decomposition must include the following for each irreducible quadratic: ax^{2} + bx + c to the m.0511
Once again, this is the same thing of stepping out m times.0516
A_{1}x + B_{1} over ax^{2} + bx + c: notice how, previously, it was A over some x plus a constant,0520
because we had that the degree of the top was one below the degree on the bottom.0528
So, the degree on the top here is a linear factor on top, because we have a quadratic factor on the bottom.0532
Previously, we had a constant factor on top, because we had a linear factor on the bottom.0537
So, A_{1}x + B_{1} + A_{2}x + B_{2}, over...that should be a capital...0541
ax^{2} + bx + c, now squared, because we are doing our second step...0548
we do this out m steps, until we are at our m constants, and we are at the m^{th} exponent on the bottom.0552
So, A_{1}...all of these capital letters...B_{1} to B_{m}...they are all just constants.0558
And we will figure them out later.0564
For example, if we had (x^{2}  4x + 5)^{2} on our bottom, then we have some stuff0565
it doesn't really matter right nowx^{2}  4x + 5, squared; so we are going to have this happen twice.0571
The first one will be just to the one; the second time, it will be squared.0576
And on the top, it is going to be Ax + B and Cx + D.0580
So, A, B, C, D...these are all just constants; we will deal with them later.0585
If we have multiple irreducible quadratics, we just do each of these.0590
For example, if we have x^{2} + 2x + 3, that is an irreducible; and x^{2} + 4, squared...0593
then we have Ax + B, x^{2} + 2x + 3...there is only one of them, so it only shows up once,0599
+ Cx + D over x^{2} + 4...it is going to show up twice because of that 2, so it shows up a second time here.0605
And we have Ex + F; so we just keep doing this process with capital letter, x, + capital letter, plus capital letter, x, plus capital letter,0612
until we have worked out all of the times that we had things showing upa total of 3, because we had one there and two there.0620
So, we have a total of 3.0627
If we have both types, a linear and an irreducible quadratic, mixed together in the denominator, that is OKperfectly fine.0629
We just decompose based on both of the rules.0634
So, for example, if we had (x + 9)^{2}, then the x + 9's are going to follow this A, this singleconstant format,0637
because linears just had one constant on the top; x + 9, and then (x + 9)^{2}.0645
And then, we switch to the other rule, x^{2} + 1; now we are dealing with an irreducible quadratic.0650
So, x^{2} + 1...and we have Cx + D on top, because we have to be able to have...0656
since we have a quadratic on the bottom, we now have linear factors on the top.0661
When we are dealing with linear factors on the bottom, even if it is multiple linear factors, it is just constants on the top.0665
All right, now, finally: how do we figure out what those numerators are?0671
We have mentioned that the numerators are built out of constants.0674
Remember: A for linear factors (constants for linear factors); Bx + C...linear factors for irreducible quadratic factors.0677
But we haven't talked about how we actually find out what these capital letters' values are.0685
We solve for the constants by doing partial fraction decomposition.0690
We set them up in this format, and then we multiply each side of the equation by the original denominator.0693
It will make more sense as we work through examples; let's look at this example.0698
If we have (3x^{2} + 3x  4)/(x + 3), one factor, times (x^{2}  x + 2), an irreducible quadratic factor,0701
then we would be able to break it into A/(x + 3) (we get this right here)...and x^{2}  x + 2 would be Bx + C/(x^{2}  x + 2).0709
So, we know, from what we just talked about (this process that we just went through) that that is how it breaks up.0721
But how do we figure out what A and Bx + C are?0726
Well, notice: we can just work things out the way we would a normal algebra thing.0728
We multiply both sides, because we want to get rid of this denominator.0734
We will multiply by x + 3 on the left side, and x^{2}  x + 2 on both sides.0737
So, we do the same thing on this side, as well: (x + 3)...we will colorcode this...and (x^{2}  x + 2).0746
So, the (x + 3) will distribute to the one on the left and the one on the right.0759
It has to hit both of them, because it is distributed, because there is this plus sign right here.0764
So, (x + 3) distributes to both of them; similarly, (x^{2}  x + 2) will distribute to both of these, as well.0768
Now, notice: on the left side, we have (x + 3) on the bottom, and we are multiplying by (x + 3); so the (x + 3)'s cancel out.0776
(x^{2}  x + 2) is right here and here; they cancel out.0783
What about on the left side? Well, we have (x + 3) and (x + 3); so here, it is going to cancel this and this.0788
But it will still have the (x^{2}  x + 2) coming through.0796
What about the Bx + C over (x^{2}  x + 2)?0799
Well, the (x + 3) is still going to come throughit doesn't cancel out any factors here.0801
But (x^{2}  x + 2) is the same thing, so it will cancel out that part of the factor coming in.0805
So, we will see that we have A times (x^{2}  x + 2) and Bx + C times (x + 3) coming through.0809
This part right here manages to come through on the Bx + C, and this part right here manages to come through on the A.0819
And on the left, since we canceled out everything on the bottom, we just have what we had on our numerator originally.0828
That just makes its way down; so we have 3x^{2} + 3x  4 = A times x^{2}  x + 2, plus Bx + C times (x + 3).0834
Which...we can then expand everything on the right, and we can work things out by creating a system of linear equations.0845
The examples are going to help greatly in clarifying how this works, so check them out,0851
because we will actually see how this process of setting up our system of linear equations works,0855
and solving for what these values are, in each of these examples.0859
Not Example 2, but we will see it in a lot of the examples.0863
And it will make a whole lot more sense as we see it in practice.0866
So, let's go check it out: the first example is a nice, simple example to get things started with.0868
Find the partial fraction decomposition for (4x + 3)/x(x + 3).0873
So, we have 4x + 3, over x(x + 3); that is going to be equal to...we have two linear factors on the bottom, so A/x, plus our other linear factor, B/(x + 3).0878
Now, at this point, we can multiply everything by x and (x + 3).0900
Great; so, on the left side, that is just going to cancel this stuff out; and on the right side, it will end up distributing.0912
So, on the left side, we have 4x + 3 = A...now, notice: we have the x multiplying here; that is going to cancel out.0917
So, the x's parts get canceled out; but we are left with x + 3 multiplying on it.0926
Plus B times (x + 3)...so the x + 3 part cancels out; I will switch colors here.0931
x gets through; but the (x + 3) cancels out the denominator; greatwe have managed to get rid of all of our denominators.0939
That is great; it will make things easier to see what is going on.0946
And we have some stuff multiplying through; so at this point, 4x + 3 on the left equals Ax + 3A + Bx.0949
Now, notice: we see here that we have a 3 constant.0965
Now, what constants do we have on the right side?0970
We have to have all of our x's match up and all of our constants match up.0972
Well, the only thing that is actually a constant on the right side is this 3A, because we have Ax and Bx.0976
But we don't have any B just as B; we don't have any constants of B.0982
So, it must be that 3 and 3A are equal, because they are all of the constants that show up on either side of our equation.0986
The same thing has to be on the left and the right; otherwise, it is not an equation.0992
That tells us that 3 = 3A; so what is our A?0996
We divide both sides by 3; we get 1 = A; so now we have figured out what A is.1001
What about the other part? Well, we have that 4x; 4x is equal to...let's give it a special thing; we will give it a curly around...1007
4x is equal to all of our x's on the right side, put together.1019
So, that is our Ax, combined with our Bx; so it must be the case that 4x is equal to Ax + Bx,1023
because the same number of x's must be on the right side as on the left side.1034
So, 4x must equal Ax + Bx, because these are our sources of x on the right side.1039
4x = Ax + Bx; we just figured out what A is, so we can plug that in over here; so we have 4x = Ax + Bx...1045
Now, notice...actually, before we do that, even, we have 4x = Ax + Bx; so if that is the case, let's just get rid of these x's.1055
All right, we know that it must be the case that, since 4x = Ax + Bx...well, that has to be true for any x that we plug in, at all.1063
So, it must be the case that 4 = A + B; we can divide the x out on both sides, and we cancel that to 4 = A + B.1072
Now, we can substitute in that 1; that will make it even clearer what is going on: 4 = 1 + B.1082
Subtract the 1; we get 3 = B; so at this point, we have figured out what A is; we have figured out what B is.1088
So, we see that we can now go back to our partial fraction decomposition.1094
We can plug in actual numbers, and we can have what that is equivalent to; and we have 1 over (x + 3) over (x + 3).1098
And there is our answer; great.1111
All right, the next one: Write out the form of the partial fraction decomposition, but do not solve for the constant.1114
That is good, because this would be really hard, if we actually had to solve it.1120
So, first we have x, and then (x  7)^{2}, and then 2x^{2} + 1, an irreducible quadratic, cubed.1123
All right, we have A/x, plus...the next one is a linear, so it is also just one constant up top, (x  7);1131
but it is squared in our original thing, so it gets squared; so another constant...C/(x  7)...this time,1142
it is two of them; it is squared; plus Dx + E, because now we are dealing with an irreducible quadratic.1149
The first time it showed up, so 2x^{2} + 1, plus Fx + G...just keep going with letters;1158
I am just counting off the alphabet at this point, Fx + G...A, B, C, D, E, F, G...I don't want to make a mistake with my alphabet!...1166
2x^{2} + 1...now we are at the second time, so squared...Hx + I, over (2x^{2} + 1)^{3}.1173
And that is the partial fraction decomposition.1186
We haven't figured out what A, B, C, D, E, F, G, H, I are.1187
But we could, if we multiplied, because we know that this thing here is equal to this thing here, once we get those correct constants in.1193
So, we could multiply the left side and the right side by that denominator.1202
And everything would cancel out, and we would be left with this awful, awful massive thing.1206
But we could solve it out, slowly but surely; luckily, all it asked for was to just set things up1209
set up the partial fraction decomposition, but not solve for these constants.1215
So, happily, we can just see that this is how this breaks down.1219
We break it down this way; we use each of the rules, based on whether it is linear or it is an irreducible quadratic.1222
And they show up the number of times of each of their exponents.1227
Great; the third example: Find the partial fraction decomposition for (x^{5} + 3x^{3} + 7x)/(x^{4} + 4x^{2} + 4).1232
The first thing to notice is that it has a degree 5 on top and a degree 4 on the bottom, so we start as improper.1241
So, since it is improper, we have to use polynomial division to break it into a format that is proper, first.1248
How many times does x^{4} + 4x^{2} + 4 go into this thing here?1254
So, let's set up our polynomial division: 4x^{2} + 4 goes into x^{5} +...we have no x^{4}'s...1258
plus 3x^{3}...we have no x^{2}'s...+ 7x; and we have no constants.1270
OK, how many times does x^{4} go into x^{5}?1279
It goes in x, because x times x^{4} gets us x^{5}; and we do it on the other things...+ 4x^{3} + 4x.1281
So, we subtract all of this; let's distribute that subtraction...1294
x^{5}  x^{5} becomes 0no surprise there.1299
3x^{3}  4x^{3} becomes x^{3}; 7x  4x becomes positive 3x.1303
We can now bring down everything else, but as soon as we do that, we realize that there is nothing else to be done here,1311
because how many times does x^{4} go into 0x^{4}? It goes in 0 times.1324
It can't fit in at all, because it is just a 0 to begin with.1329
So, we found our remainder; our remainder is what remains: our x^{3} here, our 3x, and that is it.1331
x^{3} + 3x is what remains.1338
That means that what we originally started with, x^{5} + 3x^{3} + 7x,1342
all over x^{4} + 4x^{2} + 4, is equal to...x came out of it, so x plus what our remainder was,1350
x^{3} + 3x, over that original denominator, x^{4} + 4x^{2} + 4; great.1366
We need some more room; let's flip to the next page.1377
That is what we figured out: we figured out that what we started with breaks into this thing right here.1380
So, now let's figure out how we can decompose this part on the right.1386
We will work on decomposing this first, but we can't forget (circle it in red, so we don't forget) that x, because that is still part of that function.1391
We can't actually decompose it without putting that back in at the very end.1400
Otherwise, we will have decomposed a different function; we will have decomposed this function right here,1403
but forgotten about what we originally started with, because what we originally started with includes this x.1408
All right, x^{3} + 3x...oh, that is one thing I forgot to do last time.1413
We had x^{4} + 4x^{2} + 4; I quietly factored that, without ever mentioning that it factors into (x^{2} + 2)^{2}.1418
I'm sorry about that; hopefully that isn't too confusing.1434
x^{3} + 3x...over (x^{2} + 2)^{2}...that is going to be equal to Ax + B,1437
because it is an irreducible quadratic, divided by x^{2} + 2, plus Cx + D, over (x^{2} + 2)^{2}.1446
At this point, we multiply both sides by (x^{2} + 2)^{2}, (x^{2} + 2)^{2}, (x^{2} + 2)^{2}.1458
So, that cancels out here, and we are left with x^{3} + 3x =...what happens to the Ax + B?1470
Well, its denominator gets canceled, but then it is still left getting hit with one more x^{2} + 2.1478
What about Cx + D? Well, its denominator just gets canceled; there is nothing left after (x^{2} + 2)^{2} hits,1485
because its denominator was the same thing, so we are just left with Cx + D.1492
So, let's expand that, and let's also put this in a different color, just so we don't get confused by the separation.1495
I am not sure how much that will help us.1503
3x =...Ax times x^{2} becomes Ax^{3}; Ax times 2 becomes 2Ax.1507
Bx^{2} + 2B...and we will also bring along + Cx + D.1515
At this point, we can set things up so we can see it a little more clearly by putting everything together that comes in a single form.1524
We have Ax^{3}, and we will leave a blank for the x^{2}'s...+ 2Ax.1534
And then, we can write over here + Bx^{2} + 2B...oops, sorry; that is going to have to leave a blank there, as well...1541
+ 2B...and then + Cx, and then + D; notice how this works.1550
We have our cubes here (degree 3 things here); degree 2 things here...1559
well, not quite "degree," because they are not the whole polynomial,1566
but things with exponent 3 here, exponent 2 here, exponent 1 here, and exponent 0 here.1568
So, now we can compare that over here, where we have exponent 3 here and exponent 1 here.1573
So, all of the things that are going to connect to the x^{3}, which we can also see as 1 times x^{3}, are just Ax^{3} here.1578
So, it must be that 1 is equal to A; otherwise, we wouldn't have x^{3} in the end.1587
We now know that 1 = A; great.1593
What about 2Bx^{2}? Well, we have 0x^{2}; we know that 0x^{2} = Bx^{2}.1598
Things of degree 2...well, we could have just written that as 0 = B, because our things of degree 2 have to line up with that number B.1608
So, 0 = B; what about our D here? We can figure out our D.1615
Well, we have a constant...+ 0 here; so it must be that 0 is equal to 2B + D.1622
Well, we already know that B equals 0 here; so that just knocks out; so we are now told that 0 equals D, as well.1632
Finally, we are left figuring out what C is; so we know that 3, here, equals 2A, because there are that many x,1641
because we have 3x and 2Ax and Cx, so it must be that 3 = 2A + C, when we combine them all together.1650
We plug in numbers: 3 = 2 times...what is our A? It is 1, so 1, plus C; so we get 3 = 2 + C, or we add 2 to both sides: 5 = C.1662
All right, so 5 = C; so at this point, we have managed to figure out everything that goes into our decomposition.1684
So, we can write that decomposition out: Ax + B...we have 1 as our A, so x...what is our B?1692
Our B is 0, so it just isn't anything; x^{2} + 2, plus Cx + D...our D was 0; our C was 5.1701
So, it is going to be 5x/(x^{2} + 2)^{2}.1712
And we can't forget that x that was originally there; so x plus that...that is our entire partial fraction decomposition.1719
It breaks down into that one right there.1728
All right, the final example: this one is a rough one, but this is the absolute hardest you would end up coming across in class and a test.1731
So, don't worry: this is basically the top difficulty you will have to deal with, probably.1741
You might see something a little bit harder, but this is really about as hard as it is going to get.1746
All right, so happily, they already factored it for us, and we have (t  2)^{2}(t^{2} + 2).1750
So, this is degree 2; this is also degree 2; so that combines to degree 4 on the bottom, degree 3 on the top.1756
So, we are proper; our bottom is already factored; we just need to actually work through the decomposition now.1762
We have t^{3} + 8t^{2}  6t, over (t  2)^{2}(t^{2} + 2).1769
How does that decompose? A over our linear factor, t  2, plus B over our second version of that linear factor,1781
(t  2)^{2}, plus Cx + D over our irreducible quadratic of t^{2} + 2; great.1792
At this point, we multiply both sides by the denominator, and we will get (t^{3} + 8t^{2}  6t) =1801
A times...it was hit with (t  2)^{2} + (t^{2} + 2), so it is going to cancel out one of those (t  2)'s,1814
and it will be left being multiplied by (t  2) and (t^{2} + 2).1822
Plus...B: (t  2)^{2} was in its denominator, so it is going to cancel out all of the (t  2)^{2} that hit it, but not the t^{2} + 2 at all.1828
So, we have t^{2} + 2.1838
And then finally, plus Cx + D, and it is going to have to go in parentheses, because it was hit by another entire thing.1842
And it is going to get (t  2)^{2}, but it did cancel out the (t^{2} + 2) that hit it, because it had that in its denominator.1849
At this point, we want to start simplifying things out.1856
We have A times...what is t  2 times t^{2} + 2?1859
We get t^{3}...t times 2 becomes 2t...minus 2t^{2}, minus 4,1863
plus Bt^{2} + 2B + (Cx + D) times (t  2)^{2}, which becomes t^{2}  4t + 4.1877
All right, move a little bit to the left, so we have more room; it is still the same thing on the left side.1893
We have At^{3}  2At^{2} + 2At  4A (oops, sorryI just switched colors there)1899
plus Bt^{2} + 2B...break onto a new line now, so I can see what is going on...1920
Cx times t^{2} will become C...oops, I made a mistake right from the beginning.1926
It should not be x; it should be (I just stuck with what I have been used to)...the variable we are dealing with isn't x; it is t here.1932
So, that should have been a t the whole time; sorry about that, but that is the sort of mistake you want to catch on your end, too,1942
because we are dealing with t as our variable.1951
So, while the form was with x, as before, that is because the variable was x before.1955
But now, we are dealing with t as a variable, so the form needs to switch to using t.1958
So, let's work this out: Ct times t^{2} becomes Ct^{3}; Ct times 4t becomes 4Ct^{2};1963
Ct times +4 is + 4Ct; + Dt^{2}  4Dt + 4D; OK, we have a lot of stuff here.1973
So, we can say that this is (t^{3} + 8t^{2}  6t) =1985
and we can put this in that form, again, of thing with exponent 3, exponent 2, exponent 1, exponent constant;1993
minus 2At^{2} + 2At  4A; the next one is + Bt^{2} + 2B + Ct^{3}  4Ct^{2} + 4Ct,2000
and then + Dt^{2}  4Dt + 4D.2027
So, from this, we will be able to figure out that all of our t^{3}'s, our line of t^{3}'s here,2036
has to be the same as our line of t^{3}'s here.2042
So, we will be able to get things like 1 = A + C.2044
Now, this is a lot of stuff to have to work through; we have 4 different variables;2048
we are going to end up having different simultaneous linear equations that we are going to have to solve through.2053
If you have done much work with simultaneous linear equations, you know that that is going to be kind of a pain to work through.2057
So, at this point, we might think, "Oh, I'm lazy; is there anything clever I can do?"2062
Is there a clever way to work through thissome little trick I could use?2067
Well, if we look back to what we originally had when we multiplied out that denominator,2071
we might think, "Oh, look, there is a t  2 here; there is a t  2 here; but there is an absence of t  2 on our B."2076
So, if we realize that, we could say, "Well, if we plug in t = 2, that would cause..."2086
now, none of this stuff up here is going to be true; but we are going to plug in t = 2;2093
and we are going to realize that, if we plug in t = 2, that causes this to turn to 0, which knocks out our A terms.2099
And it will cause this here to turn to 0, as well, and also knock out our C and our D terms.2105
And we will be left with just B times t^{2} + 2, where t is 2only t is 22110
and t^{3} + 8t^{2}  6t, when t is 2.2117
Now, this is true for any t; so that means we can plug in any t we want,2122
and if there is a convenient way to get certain things to disappear by plugging in a cleverly chosen t, it is fair.2127
Anything we can do to help us work through thisanything that makes it easieris fair.2134
So, in this case, we notice that by plugging in this carefullychosen t, we can get certain things to disappear.2138
Now, it is true for any t; this equation, this whole thing here, is true in general for any t, before we cross stuff out.2144
But in the specific case where we plug in t = 2, certain things will cancel out, and we will be able to figure things out in a very easy way mathematically.2152
So, we plug this in; and we know that, on our left side, we are going to have 2^{3} + 8(2)^{2}  6(2).2161
That equals (sorry, there is not quite a lot of room here; I am going to draw in a line, just so we don't get confused)...2173
That equals B times 2^{2} + 2; OK.2180
So, we have 2^{2} is 8, plus 8(2)^{2} is 4, so 32, minus 6 times 2 is 12, so  12, equals B times 4 + 2, or 6.2186
Simplify the left side: 8  12 is 20; 32  20 gets us 12, equals 6B; we divide both sides by 6, and we get 2 = B.2208
So, by use of some cleverness, we were able to skip having to do four simultaneous linear equations,2221
to where eventually we are effectively going to bring it down to 3.2227
And we will be able to not have to get confused by this most complicated column, where we have four different things all going together.2230
We can just be done with B; B is already figured out, which will be really helpful and make things easier on us later on.2236
We could have solved out those four simultaneous linear equations, because each one will produce things.2242
Remember how the first column produced 1 = A + C; each one of these columns will produce an equation.2247
And from four simultaneous linear equations, solving for four variables, we will be able to do it; it is just kind of a pain.2253
So, we came up with this clever way, and we were able to figure out 2 = B by just happenstance.2259
If we plug in t = 2, it made everything but the B terms disappear, and that just left us with a single equation that was pretty easy to solve.2263
And we figured out that 2 = B; all right, well, let's work through it now.2271
So, we figured out B = 2, and now we have t^{3} + 8t^{2}  6t =2275
At^{3} + Ct^{3}  2At^{2} + Bt^{2}  4Ct^{2} + Dt^{2} + 2At + 4Ct  4Dt  4A + 2B + 4D; wow!2282
OK, so at this point, let's put columns to columns; so t^{3} goes to our At^{3} + Ct^{3}.2294
So, it must be that 1 is equal to A + C.2301
We can do this column here; I accidentally cut through that negative signhopefully that wasn't too confusing.2309
So, that one here lines up with the 6t; so we have that 6 (let's write it on a separate location) equals 2A + 4C  4D.2316
They all have t's showing up, so we just divide out the t's, and we are left with this simple linear equation that we can work with.2334
And then, finally, the last one of constants: what is our constant? Our constant is 0.2340
So, we have 0 = 4A + 2B + 4D; for extra credit, let's just see what this would have been.2347
How many t^{2}'s did we have? We had 8, so in our difficulttoread yellow (I'll make it black,2357
just because black is easier to read), 8 = 2A + B  4C + D.2362
We could have worked that one out, but I wanted you to see what it would have been.2371
But we will actually end up having enough information, because we have three equations and three unknowns;2375
so we have enough with the red, blue, and green things.2379
So, at this point, let's figure out A first; so that means we need to solve for everything that isn't A,2382
so that we can plug them in, get rid of everything else, and just have A.2389
So, on the left side, we have 1 = A + C; so we solve for C, and we get A  1 = C.2392
OK; on the green one, we plug in what we know for our B.2405
We have 0 = 4A + 2(2 + 4D); 0 = 4A + 4 + 4D; 0 =...let's divide everything by 4, just to make things easier...A + 1 + D.2409
So, we move that stuff over, and we get: add A on both sides; subtract by 1 on both sides; A  1 = D.2430
So, we have A  1 = D; we have A  1 = C; so we can plug this information into this equation, and we will be able to get something that is just using A.2439
6 = 2(A)...that actually stays around, because it was just A from the beginning; 4...what did we figure out C was?2460
C is the same thing as A  1, minus 4 times...what did we figure out D is?2468
It is the same thing as A  1; simplify that out: 6 = 2A + 4(A), so we will make that  4A, minus 4...2475
4A  4A...and 4 times 1 becomes + 4; so at this point, we see that we have 4 and +4, so those cancel each other out.2489
We have 6 = 2A  8A; 6 = 6A; and thus 1 (dividing both sides by 6) = A.2500
With 1 = A in place, we can now easily go back and figure out everything else.2513
We have A  1, so negative...plug it in, knowing that it is 1...minus 1, equals C; so we have 2 = C; great.2518
And also, plug in over here; and we have that 1  1 = D; it turns out that D is just 0.2533
So, at this point, we have figured out all of the things that we need.2545
We have this set up as (t  2) + (t  2)^{2} + (t^{2} + 2).2549
It was in the format t^{3} + 8t^{2}  6t, all over (t  2)^{2}(t^{2} + 2).2557
OK, so that is the case; we have that as something over (t  2).2579
It was A/(t  2) + B/(t  2)^{2} + (Cx + D)/(t^{2} + 2).2586
OK, at this point, we can actually plug in our numbers; we have that A is 1, so we have 1/(t  2).2601
B is 2, so we have 2/(t  2)^{2}, plus...C was 2, so 2x, and 0 was D, so that part just disappears.2608
So, we have 2x/(t^{2} + 2); oops, once again, I did it again; I got to thinking in terms of x, as opposed to t.2624
But we are using a different variable: 2t/(t^{2} + 2).2633
And there we are; we have decomposed it into its partial fractions.2638
That is pretty long; it is pretty complicated; but that is pretty much as hard as it gets.2643
As long as you break it down, and then you multiply everything out carefully,2647
and you are careful with all of your algebra and your arithmetic, you can get it into this form right here,2651
where you have this big block compared to these original things, and then you are able to figure out all of these linear equations.2656
And you can possibly be clever and figure out some way to figure out B = 2.2662
But you also can just work through a bunch of linear equations, solving for one thing at a time in terms of the other,2667
plugging them all together, and then solving it out.2673
Sometimes it takes a lot of work; sometimes it goes kind of slowly.2675
But ultimately, if you just keep working at it, you can get it.2678
And it is really, really useful in calculus.2680
I knowI know that you aren't going to be using it immediately.2683
But honestly, this thing makes a problem that would be, otherwise, totally impossible to do, just really easy.2685
So, it is a really handy thing in calculus.2691
All right, we will see you at Educator.com latergoodbye!2693
2 answers
Last reply by: Peter Ke
Tue Nov 10, 2015 8:11 PM
Post by Peter Ke on November 8, 2015
At 11:40 "Quick Example", I understand at the left side of the equation, how you got rid of the denominator by multiplying but then how you got the right side though?
The right side was A(x^2  x + 2) + (Bx+C)(x+3). How did you get this!?
I really don't get it...
1 answer
Last reply by: Professor SelhorstJones
Tue Sep 9, 2014 9:52 AM
Post by David Llewellyn on September 8, 2014
Your explanation that an expression with all exponents of x <=0 does not qualify as a polynomial (see my question in Polynomials lecture) explains why you stop the polynomial division with a remainder once you hit a constant in this lecture.
BTW, presumably, you can also have irreducible factors of any even power (x^2n) n>= 1 with 2n terms in the denominator. I can see that all polynomials will have, at least, one real factor as this polynomial will generate an odd function.