WEBVTT mathematics/math-analysis/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about partial fractions.
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Consider if we had two quotients of polynomials that we wanted to add together, like, say, 7/(x - 5) and 3/(x + 2).
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We could put them over a common denominator, then combine them; we have x - 5 here and x + 2 here,
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so we can multiply the one on the left by x + 2, so we get (x - 5)(x + 2) on the bottom.
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Its top will get also multiplied by that (x + 2).
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For the other one, we will have (x - 5) multiplied on it, so it has (x - 5) multiplied on the top,
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so we have (x - 5)(x + 2)--a common denominator--on the bottom now.
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We multiply this out; we get 7x + 14 and 3x - 15; we combine those, and we get 10x - 1 on top.
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divided by (x - 5)(x + 2), expanded to x² - 3x - 10.
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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.
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But if we wanted to do the reverse process--what if we started with a fraction involving large polynomials,
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and we wanted to break it into smaller fractions made of the polynomials' factors?
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If we wanted to do this process in reverse, if we started at (10x - 1)/(x² - 3x - 10),
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and we somehow wanted to be able to get that into 7/(x - 5) and 3/(x + 2), what process could we go through?
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We call the smaller fractions on the right **partial fractions**, because they are parts of that larger fraction.
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So, each of these here (7/(x - 5) and 3/(x + 2))--they are each called a partial fraction.
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And the process to break it up is called **partial fraction decomposition**.
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So, whatever this method is that we haven't explored yet, to get from that larger polynomial quotient
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into these smaller partial fractions, is called partial fraction decomposition.
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So, how do we do it? To understand this lesson--to understand how we do it--you will need some familiarity with solving systems of linear equations.
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Previous experience from past algebra classes will probably be enough to get through this.
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But if you want a refresher, if you are a little confused by how this stuff is working later on,
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check out the lesson Systems of Linear Equations--that will help explain this stuff if it confuses you right now.
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You also need to know how to factor polynomials, and you will need to understand them in general,
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along with the ability to do polynomial division; you will need all of that for some of the stuff in here, as well.
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Now, sadly, we won't be able to see the application of partial fraction decomposition in this course.
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However, it is quite useful in calculus, where it will allow us to solve otherwise impossible problems.
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Partial fraction decomposition is this really useful thing that lets us break up these complicated things
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that we couldn't solve, and turn them into a form that is actually pretty easy to solve.
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It is really handy in calculus.
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We won't be able to see its use in this course; we won't see it any time soon.
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But it is helpful to practice it now, just like we practiced stuff about factoring complicated polynomials in algebra,
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before we really had a great understanding of what it was all about.
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We are practicing something so that we can use it later on.
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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.
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All right, let's get to actually figuring out how to do this.
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If we have a polynomial fraction in the form numerator polynomial divided by denominator polynomial,
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a normal rational function format, there are two possibilities in regards to the degrees of the top and bottom polynomials.
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We call them proper (proper is when the degree of the numerator polynomial is less than the degree of the denominator polynomial--
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the degree of n is less than the degree of the denominator) and improper (when the degree of our numerator
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is greater than or equal to the degree of the denominator).
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To decompose the fraction, it must be proper; we have to be in this proper format.
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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.
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So, what do we do if the fraction is improper?
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We have to turn it into a proper thing; so if the fraction is improper, and we want to decompose it,
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we have to make it proper through polynomial division.
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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.
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Remember, when you divide your polynomial division, the remainder from the division
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goes back onto our original denominator, which we will then decompose.
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We will be able to decompose the part that is the remainder.
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The other part that comes out cleanly--well, that is just going to be there.
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That is just going to be a polynomial that is then going to be added to whatever ends up coming out of our decomposition.
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We will see an example of this in Example 3.
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Once we have a proper polynomial fraction to decompose into partial fractions
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(remember, proper: our numerator has to be smaller than our denominator before we can enact this process),
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once we have done that, we can then factor the denominator.
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Our next step is to factor the denominator.
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After the denominator is broken into its smallest possible factors, we are ready to decompose.
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Now, there are two types that the smallest possible factors can come in.
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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.
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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.
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So, we will have linear factors in this; or we could have irreducible quadratic factors,
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that is, things ax² + bx + c that can't be broken up further into linear factors.
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There is no way to break them up further in the reals.
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So, ax² + bx + c...and once again, it will be raised to some power;
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there will be m of them, so it is raised to the m, because there are m of them multiplying together.
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Remember: "irreducible" means it can't be broken up further in the reals.
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That means quadratics like x² + 1 or 5x² - 3x + 20, because they have no roots.
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x² + 1 has no roots; x² + 1 = 0...there are no solutions, in the reals, at least.
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If we allow for the complex, there are; but we are not working with the complex.
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There are no solutions in the reals; so, since there are no solutions in the reals, it can't be factored any further.
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x² + 1 is irreducible, therefore.
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The next step of decomposition will behave differently, depending on which flavor of factor we have--
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whether we are at a linear factor or we are at an irreducible factor; and we will look at the two, one after another.
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Linear factors: the partial fraction decomposition must include the following for each linear factor that is in this form (ax + b)^m.
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So, it is going to have this in its decomposition: A₁/(ax + b) + A₂/(ax + b)² +...
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notice that we are doing this where it is going to keep stepping up with this exponent on the bottom,
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as we keep putting these things in, until eventually we get up to our mth step,
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because we had m of them to begin with, so we have to have m of them in the end.
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And each one of these A's on top (A₁ up until A<font size="-6">m</font>) are all just constant real numbers.
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We are using A with a subscript, this A₁, A₂, A₃ business,
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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.
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Anyway, the point is that they are all going to be constant real numbers.
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For example, if we had (x + 7)³, then we have some stuff up top.
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We don't really care what the stuff is right now, because we just want to see how it will break out.
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We will deal with the stuff later.
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If we have stuff over (x + 7)³, then it is going to break into 3 of these, because of this "cubed."
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We will have A/(x + 7) + B (we can switch into just using letters in general, because we know
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that each one of these capital letters just means some constant real number; we will figure them out later--
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that will come up later--don't worry)...A/(x + 7) + B/(x + 7)² + C/(x + 7)³.
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Notice: we started cubed, and so we have three of these; we step up each time with the exponent,
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until we have gotten to the number of our multiplicity that we originally started with--
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whatever the exponent was originally on the factor.
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If we have multiple linear factors, we do it for each one of these; so if we have (2x + 3)², x¹
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(since there is nothing there), (x - 5)¹ (since there is nothing there)...we have A/(2x + 3) + B/(2x + 3)²
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(because remember: it was squared to begin with, so it has to have 2 of itself) + C/x.
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And there is only to the 1, so there is only going to be one of it; plus D/(x - 5).
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And it is only 1, because there is only one of them to begin with.
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And each one of these A, B, C, D...they are each just constant numbers.
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We are going to figure them out later on; don't worry.
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Irreducible quadratic factors: the method is very similar for irreducible quadratic factors.
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The decomposition must include the following for each irreducible quadratic: ax² + bx + c to the m.
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Once again, this is the same thing of stepping out m times.
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A₁x + B₁ over ax² + bx + c: notice how, previously, it was A over some x plus a constant,
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because we had that the degree of the top was one below the degree on the bottom.
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So, the degree on the top here is a linear factor on top, because we have a quadratic factor on the bottom.
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Previously, we had a constant factor on top, because we had a linear factor on the bottom.
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So, A₁x + B₁ + A₂x + B₂, over...that should be a capital...
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ax² + bx + c, now squared, because we are doing our second step...
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we do this out m steps, until we are at our m constants, and we are at the mth exponent on the bottom.
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So, A₁...all of these capital letters...B₁ to B<font size="-6">m</font>...they are all just constants.
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And we will figure them out later.
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For example, if we had (x² - 4x + 5)² on our bottom, then we have some stuff--
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it doesn't really matter right now--x² - 4x + 5, squared; so we are going to have this happen twice.
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The first one will be just to the one; the second time, it will be squared.
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And on the top, it is going to be Ax + B and Cx + D.
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So, A, B, C, D...these are all just constants; we will deal with them later.
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If we have multiple irreducible quadratics, we just do each of these.
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For example, if we have x² + 2x + 3, that is an irreducible; and x² + 4, squared...
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then we have Ax + B, x² + 2x + 3...there is only one of them, so it only shows up once,
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+ Cx + D over x² + 4...it is going to show up twice because of that 2, so it shows up a second time here.
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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,
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until we have worked out all of the times that we had things showing up--a total of 3, because we had one there and two there.
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So, we have a total of 3.
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If we have both types, a linear and an irreducible quadratic, mixed together in the denominator, that is OK--perfectly fine.
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We just decompose based on both of the rules.
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So, for example, if we had (x + 9)², then the x + 9's are going to follow this A, this single-constant format,
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because linears just had one constant on the top; x + 9, and then (x + 9)².
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And then, we switch to the other rule, x² + 1; now we are dealing with an irreducible quadratic.
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So, x² + 1...and we have Cx + D on top, because we have to be able to have...
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since we have a quadratic on the bottom, we now have linear factors on the top.
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When we are dealing with linear factors on the bottom, even if it is multiple linear factors, it is just constants on the top.
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All right, now, finally: how do we figure out what those numerators are?
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We have mentioned that the numerators are built out of constants.
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Remember: A for linear factors (constants for linear factors); Bx + C...linear factors for irreducible quadratic factors.
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But we haven't talked about how we actually find out what these capital letters' values are.
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We solve for the constants by doing partial fraction decomposition.
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We set them up in this format, and then we multiply each side of the equation by the original denominator.
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It will make more sense as we work through examples; let's look at this example.
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If we have (3x² + 3x - 4)/(x + 3), one factor, times (x² - x + 2), an irreducible quadratic factor,
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then we would be able to break it into A/(x + 3) (we get this right here)...and x² - x + 2 would be Bx + C/(x² - x + 2).
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So, we know, from what we just talked about (this process that we just went through) that that is how it breaks up.
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But how do we figure out what A and Bx + C are?
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Well, notice: we can just work things out the way we would a normal algebra thing.
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We multiply both sides, because we want to get rid of this denominator.
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We will multiply by x + 3 on the left side, and x² - x + 2 on both sides.
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So, we do the same thing on this side, as well: (x + 3)...we will color-code this...and (x² - x + 2).
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So, the (x + 3) will distribute to the one on the left and the one on the right.
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It has to hit both of them, because it is distributed, because there is this plus sign right here.
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So, (x + 3) distributes to both of them; similarly, (x² - x + 2) will distribute to both of these, as well.
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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.
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(x² - x + 2) is right here and here; they cancel out.
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What about on the left side? Well, we have (x + 3) and (x + 3); so here, it is going to cancel this and this.
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But it will still have the (x² - x + 2) coming through.
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What about the Bx + C over (x² - x + 2)?
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Well, the (x + 3) is still going to come through--it doesn't cancel out any factors here.
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But (x² - x + 2) is the same thing, so it will cancel out that part of the factor coming in.
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So, we will see that we have A times (x² - x + 2) and Bx + C times (x + 3) coming through.
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This part right here manages to come through on the Bx + C, and this part right here manages to come through on the A.
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And on the left, since we canceled out everything on the bottom, we just have what we had on our numerator originally.
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That just makes its way down; so we have 3x² + 3x - 4 = A times x² - x + 2, plus Bx + C times (x + 3).
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Which...we can then expand everything on the right, and we can work things out by creating a system of linear equations.
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The examples are going to help greatly in clarifying how this works, so check them out,
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because we will actually see how this process of setting up our system of linear equations works,
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and solving for what these values are, in each of these examples.
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Not Example 2, but we will see it in a lot of the examples.
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And it will make a whole lot more sense as we see it in practice.
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So, let's go check it out: the first example is a nice, simple example to get things started with.
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Find the partial fraction decomposition for (4x + 3)/x(x + 3).
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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).
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Now, at this point, we can multiply everything by x and (x + 3).
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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.
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So, on the left side, we have 4x + 3 = A...now, notice: we have the x multiplying here; that is going to cancel out.
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So, the x's parts get canceled out; but we are left with x + 3 multiplying on it.
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Plus B times (x + 3)...so the x + 3 part cancels out; I will switch colors here.
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x gets through; but the (x + 3) cancels out the denominator; great--we have managed to get rid of all of our denominators.
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That is great; it will make things easier to see what is going on.
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And we have some stuff multiplying through; so at this point, 4x + 3 on the left equals Ax + 3A + Bx.
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Now, notice: we see here that we have a 3 constant.
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Now, what constants do we have on the right side?
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We have to have all of our x's match up and all of our constants match up.
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Well, the only thing that is actually a constant on the right side is this 3A, because we have Ax and Bx.
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But we don't have any B just as B; we don't have any constants of B.
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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.
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The same thing has to be on the left and the right; otherwise, it is not an equation.
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That tells us that 3 = 3A; so what is our A?
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We divide both sides by 3; we get 1 = A; so now we have figured out what A is.
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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...
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4x is equal to all of our x's on the right side, put together.
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So, that is our Ax, combined with our Bx; so it must be the case that 4x is equal to Ax + Bx,
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because the same number of x's must be on the right side as on the left side.
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So, 4x must equal Ax + Bx, because these are our sources of x on the right side.
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4x = Ax + Bx; we just figured out what A is, so we can plug that in over here; so we have 4x = Ax + Bx...
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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.
00:17:43.500 --> 00:17:51.900
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.
00:17:51.900 --> 00:18:02.200
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.
00:18:02.200 --> 00:18:08.400
Now, we can substitute in that 1; that will make it even clearer what is going on: 4 = 1 + B.
00:18:08.400 --> 00:18:14.500
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.
00:18:14.500 --> 00:18:17.900
So, we see that we can now go back to our partial fraction decomposition.
00:18:17.900 --> 00:18:31.400
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).
00:18:31.400 --> 00:18:33.800
And there is our answer; great.
00:18:33.800 --> 00:18:40.300
All right, the next one: Write out the form of the partial fraction decomposition, but do not solve for the constant.
00:18:40.300 --> 00:18:43.600
That is good, because this would be really hard, if we actually had to solve it.
00:18:43.600 --> 00:18:51.200
So, first we have x, and then (x - 7)², and then 2x² + 1, an irreducible quadratic, cubed.
00:18:51.200 --> 00:19:02.500
All right, we have A/x, plus...the next one is a linear, so it is also just one constant up top, (x - 7);
00:19:02.500 --> 00:19:09.100
but it is squared in our original thing, so it gets squared; so another constant...C/(x - 7)...this time,
00:19:09.100 --> 00:19:18.600
it is two of them; it is squared; plus Dx + E, because now we are dealing with an irreducible quadratic.
00:19:18.600 --> 00:19:26.500
The first time it showed up, so 2x² + 1, plus Fx + G...just keep going with letters;
00:19:26.500 --> 00:19:32.900
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!...
00:19:32.900 --> 00:19:45.700
2x² + 1...now we are at the second time, so squared...Hx + I, over (2x² + 1)³.
00:19:45.700 --> 00:19:47.600
And that is the partial fraction decomposition.
00:19:47.600 --> 00:19:53.200
We haven't figured out what A, B, C, D, E, F, G, H, I are.
00:19:53.200 --> 00:20:02.400
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.
00:20:02.400 --> 00:20:05.700
So, we could multiply the left side and the right side by that denominator.
00:20:05.700 --> 00:20:09.500
And everything would cancel out, and we would be left with this awful, awful massive thing.
00:20:09.500 --> 00:20:15.300
But we could solve it out, slowly but surely; luckily, all it asked for was to just set things up--
00:20:15.300 --> 00:20:19.100
set up the partial fraction decomposition, but not solve for these constants.
00:20:19.100 --> 00:20:21.900
So, happily, we can just see that this is how this breaks down.
00:20:21.900 --> 00:20:27.600
We break it down this way; we use each of the rules, based on whether it is linear or it is an irreducible quadratic.
00:20:27.600 --> 00:20:32.400
And they show up the number of times of each of their exponents.
00:20:32.400 --> 00:20:40.700
Great; the third example: Find the partial fraction decomposition for (x⁵ + 3x³ + 7x)/(x⁴ + 4x² + 4).
00:20:40.700 --> 00:20:48.500
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.
00:20:48.500 --> 00:20:54.200
So, since it is improper, we have to use polynomial division to break it into a format that is proper, first.
00:20:54.200 --> 00:20:58.200
How many times does x⁴ + 4x² + 4 go into this thing here?
00:20:58.200 --> 00:21:10.600
So, let's set up our polynomial division: 4x² + 4 goes into x⁵ +...we have no x⁴'s...
00:21:10.600 --> 00:21:18.700
plus 3x³...we have no x²'s...+ 7x; and we have no constants.
00:21:18.700 --> 00:21:21.200
OK, how many times does x⁴ go into x⁵?
00:21:21.200 --> 00:21:33.700
It goes in x, because x times x⁴ gets us x⁵; and we do it on the other things...+ 4x³ + 4x.
00:21:33.700 --> 00:21:39.600
So, we subtract all of this; let's distribute that subtraction...
00:21:39.600 --> 00:21:42.800
x⁵ - x⁵ becomes 0--no surprise there.
00:21:42.800 --> 00:21:51.500
3x³ - 4x³ becomes -x³; 7x - 4x becomes positive 3x.
00:21:51.500 --> 00:22:03.800
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,
00:22:03.800 --> 00:22:09.100
because how many times does x⁴ go into 0x⁴? It goes in 0 times.
00:22:09.100 --> 00:22:11.200
It can't fit in at all, because it is just a 0 to begin with.
00:22:11.200 --> 00:22:17.700
So, we found our remainder; our remainder is what remains: our -x³ here, our 3x, and that is it.
00:22:17.700 --> 00:22:22.200
-x³ + 3x is what remains.
00:22:22.200 --> 00:22:30.400
That means that what we originally started with, x⁵ + 3x³ + 7x,
00:22:30.400 --> 00:22:45.700
all over x⁴ + 4x² + 4, is equal to...x came out of it, so x plus what our remainder was,
00:22:45.700 --> 00:22:57.300
-x³ + 3x, over that original denominator, x⁴ + 4x² + 4; great.
00:22:57.300 --> 00:23:00.000
We need some more room; let's flip to the next page.
00:23:00.000 --> 00:23:06.400
That is what we figured out: we figured out that what we started with breaks into this thing right here.
00:23:06.400 --> 00:23:10.800
So, now let's figure out how we can decompose this part on the right.
00:23:10.800 --> 00:23:19.700
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.
00:23:19.700 --> 00:23:23.600
We can't actually decompose it without putting that back in at the very end.
00:23:23.600 --> 00:23:27.900
Otherwise, we will have decomposed a different function; we will have decomposed this function right here,
00:23:27.900 --> 00:23:33.100
but forgotten about what we originally started with, because what we originally started with includes this x.
00:23:33.100 --> 00:23:38.600
All right, -x³ + 3x...oh, that is one thing I forgot to do last time.
00:23:38.600 --> 00:23:53.800
We had x⁴ + 4x² + 4; I quietly factored that, without ever mentioning that it factors into (x² + 2)².
00:23:53.800 --> 00:23:57.300
I'm sorry about that; hopefully that isn't too confusing.
00:23:57.300 --> 00:24:06.200
-x³ + 3x...over (x² + 2)²...that is going to be equal to Ax + B,
00:24:06.200 --> 00:24:17.900
because it is an irreducible quadratic, divided by x² + 2, plus Cx + D, over (x² + 2)².
00:24:17.900 --> 00:24:29.900
At this point, we multiply both sides by (x² + 2)², (x² + 2)², (x² + 2)².
00:24:29.900 --> 00:24:37.900
So, that cancels out here, and we are left with -x³ + 3x =...what happens to the Ax + B?
00:24:37.900 --> 00:24:45.300
Well, its denominator gets canceled, but then it is still left getting hit with one more x² + 2.
00:24:45.300 --> 00:24:51.900
What about Cx + D? Well, its denominator just gets canceled; there is nothing left after (x² + 2)² hits,
00:24:51.900 --> 00:24:55.400
because its denominator was the same thing, so we are just left with Cx + D.
00:24:55.400 --> 00:25:03.000
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.
00:25:03.000 --> 00:25:06.900
I am not sure how much that will help us.
00:25:06.900 --> 00:25:15.300
3x =...Ax times x² becomes Ax³; Ax times 2 becomes 2Ax.
00:25:15.300 --> 00:25:24.500
Bx² + 2B...and we will also bring along + Cx + D.
00:25:24.500 --> 00:25:33.900
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.
00:25:33.900 --> 00:25:41.100
We have Ax³, and we will leave a blank for the x²'s...+ 2Ax.
00:25:41.100 --> 00:25:50.000
And then, we can write over here + Bx² + 2B...oops, sorry; that is going to have to leave a blank there, as well...
00:25:50.000 --> 00:25:58.900
+ 2B...and then + Cx, and then + D; notice how this works.
00:25:58.900 --> 00:26:05.700
We have our cubes here (degree 3 things here); degree 2 things here...
00:26:05.700 --> 00:26:08.000
well, not quite "degree," because they are not the whole polynomial,
00:26:08.000 --> 00:26:13.300
but things with exponent 3 here, exponent 2 here, exponent 1 here, and exponent 0 here.
00:26:13.300 --> 00:26:18.100
So, now we can compare that over here, where we have exponent 3 here and exponent 1 here.
00:26:18.100 --> 00:26:27.300
So, all of the things that are going to connect to the -x³, which we can also see as -1 times x³, are just Ax³ here.
00:26:27.300 --> 00:26:33.400
So, it must be that -1 is equal to A; otherwise, we wouldn't have -x³ in the end.
00:26:33.400 --> 00:26:38.000
We now know that -1 = A; great.
00:26:38.000 --> 00:26:47.900
What about 2Bx²? Well, we have 0x²; we know that 0x² = Bx².
00:26:47.900 --> 00:26:55.200
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.
00:26:55.200 --> 00:27:02.600
So, 0 = B; what about our D here? We can figure out our D.
00:27:02.600 --> 00:27:12.400
Well, we have a constant...+ 0 here; so it must be that 0 is equal to 2B + D.
00:27:12.400 --> 00:27:20.700
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.
00:27:20.700 --> 00:27:30.000
Finally, we are left figuring out what C is; so we know that 3, here, equals 2A, because there are that many x,
00:27:30.000 --> 00:27:41.900
because we have 3x and 2Ax and Cx, so it must be that 3 = 2A + C, when we combine them all together.
00:27:41.900 --> 00:28:04.500
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.
00:28:04.500 --> 00:28:11.700
All right, so 5 = C; so at this point, we have managed to figure out everything that goes into our decomposition.
00:28:11.700 --> 00:28:20.700
So, we can write that decomposition out: Ax + B...we have -1 as our A, so -x...what is our B?
00:28:20.700 --> 00:28:32.300
Our B is 0, so it just isn't anything; x² + 2, plus Cx + D...our D was 0; our C was 5.
00:28:32.300 --> 00:28:39.600
So, it is going to be 5x/(x² + 2)².
00:28:39.600 --> 00:28:48.500
And we can't forget that x that was originally there; so x plus that...that is our entire partial fraction decomposition.
00:28:48.500 --> 00:28:50.900
It breaks down into that one right there.
00:28:50.900 --> 00:29:01.600
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.
00:29:01.600 --> 00:29:06.400
So, don't worry: this is basically the top difficulty you will have to deal with, probably.
00:29:06.400 --> 00:29:10.000
You might see something a little bit harder, but this is really about as hard as it is going to get.
00:29:10.000 --> 00:29:16.200
All right, so happily, they already factored it for us, and we have (t - 2)²(t² + 2).
00:29:16.200 --> 00:29:22.400
So, this is degree 2; this is also degree 2; so that combines to degree 4 on the bottom, degree 3 on the top.
00:29:22.400 --> 00:29:29.400
So, we are proper; our bottom is already factored; we just need to actually work through the decomposition now.
00:29:29.400 --> 00:29:41.600
We have -t³ + 8t² - 6t, over (t - 2)²(t² + 2).
00:29:41.600 --> 00:29:52.200
How does that decompose? A over our linear factor, t - 2, plus B over our second version of that linear factor,
00:29:52.200 --> 00:30:01.400
(t - 2)², plus Cx + D over our irreducible quadratic of t² + 2; great.
00:30:01.400 --> 00:30:14.200
At this point, we multiply both sides by the denominator, and we will get (-t³ + 8t² - 6t) =
00:30:14.200 --> 00:30:21.900
A times...it was hit with (t - 2)² + (t² + 2), so it is going to cancel out one of those (t - 2)'s,
00:30:21.900 --> 00:30:28.000
and it will be left being multiplied by (t - 2) and (t² + 2).
00:30:28.000 --> 00:30:38.300
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 t² + 2 at all.
00:30:38.300 --> 00:30:41.900
So, we have t² + 2.
00:30:41.900 --> 00:30:49.300
And then finally, plus Cx + D, and it is going to have to go in parentheses, because it was hit by another entire thing.
00:30:49.300 --> 00:30:56.600
And it is going to get (t - 2)², but it did cancel out the (t² + 2) that hit it, because it had that in its denominator.
00:30:56.600 --> 00:30:59.300
At this point, we want to start simplifying things out.
00:30:59.300 --> 00:31:03.100
We have A times...what is t - 2 times t² + 2?
00:31:03.100 --> 00:31:16.900
We get t³...t times 2 becomes 2t...minus 2t², minus 4,
00:31:16.900 --> 00:31:33.600
plus Bt² + 2B + (Cx + D) times (t - 2)², which becomes t² - 4t + 4.
00:31:33.600 --> 00:31:39.200
All right, move a little bit to the left, so we have more room; it is still the same thing on the left side.
00:31:39.200 --> 00:31:59.800
We have At³ - 2At² + 2At - 4A (oops, sorry--I just switched colors there)
00:31:59.800 --> 00:32:06.000
plus Bt² + 2B...break onto a new line now, so I can see what is going on...
00:32:06.000 --> 00:32:12.400
Cx times t² will become C...oops, I made a mistake right from the beginning.
00:32:12.400 --> 00:32:21.700
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.
00:32:21.700 --> 00:32:31.200
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,
00:32:31.200 --> 00:32:34.700
because we are dealing with t as our variable.
00:32:34.700 --> 00:32:38.400
So, while the form was with x, as before, that is because the variable was x before.
00:32:38.400 --> 00:32:42.700
But now, we are dealing with t as a variable, so the form needs to switch to using t.
00:32:42.700 --> 00:32:52.800
So, let's work this out: Ct times t² becomes Ct³; Ct times -4t becomes -4Ct²;
00:32:52.800 --> 00:33:05.300
Ct times +4 is + 4Ct; + Dt² - 4Dt + 4D; OK, we have a lot of stuff here.
00:33:05.300 --> 00:33:13.400
So, we can say that this is (-t³ + 8t² - 6t) =
00:33:13.400 --> 00:33:19.800
and we can put this in that form, again, of thing with exponent 3, exponent 2, exponent 1, exponent constant;
00:33:19.800 --> 00:33:47.400
minus 2At² + 2At - 4A; the next one is + Bt² + 2B + Ct³ - 4Ct² + 4Ct,
00:33:47.400 --> 00:33:56.500
and then + Dt² - 4Dt + 4D.
00:33:56.500 --> 00:34:01.700
So, from this, we will be able to figure out that all of our t³'s, our line of t³'s here,
00:34:01.700 --> 00:34:03.700
has to be the same as our line of t³'s here.
00:34:03.700 --> 00:34:08.400
So, we will be able to get things like -1 = A + C.
00:34:08.400 --> 00:34:13.100
Now, this is a lot of stuff to have to work through; we have 4 different variables;
00:34:13.100 --> 00:34:17.600
we are going to end up having different simultaneous linear equations that we are going to have to solve through.
00:34:17.600 --> 00:34:21.900
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.
00:34:21.900 --> 00:34:26.900
So, at this point, we might think, "Oh, I'm lazy; is there anything clever I can do?"
00:34:26.900 --> 00:34:31.000
Is there a clever way to work through this--some little trick I could use?
00:34:31.000 --> 00:34:36.000
Well, if we look back to what we originally had when we multiplied out that denominator,
00:34:36.000 --> 00:34:46.300
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."
00:34:46.300 --> 00:34:53.000
So, if we realize that, we could say, "Well, if we plug in t = 2, that would cause..."
00:34:53.000 --> 00:34:58.800
now, none of this stuff up here is going to be true; but we are going to plug in t = 2;
00:34:58.800 --> 00:35:04.700
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.
00:35:04.700 --> 00:35:10.400
And it will cause this here to turn to 0, as well, and also knock out our C and our D terms.
00:35:10.400 --> 00:35:17.000
And we will be left with just B times t² + 2, where t is 2--only t is 2--
00:35:17.000 --> 00:35:22.400
and -t³ + 8t² - 6t, when t is 2.
00:35:22.400 --> 00:35:26.900
Now, this is true for any t; so that means we can plug in any t we want,
00:35:26.900 --> 00:35:34.100
and if there is a convenient way to get certain things to disappear by plugging in a cleverly chosen t, it is fair.
00:35:34.100 --> 00:35:38.400
Anything we can do to help us work through this--anything that makes it easier--is fair.
00:35:38.400 --> 00:35:43.900
So, in this case, we notice that by plugging in this carefully-chosen t, we can get certain things to disappear.
00:35:43.900 --> 00:35:52.600
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.
00:35:52.600 --> 00:36:00.800
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.
00:36:00.800 --> 00:36:12.700
So, we plug this in; and we know that, on our left side, we are going to have -2³ + 8(2)² - 6(2).
00:36:12.700 --> 00:36:19.800
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)...
00:36:19.800 --> 00:36:25.900
That equals B times 2² + 2; OK.
00:36:25.900 --> 00:36:48.000
So, we have -2² 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.
00:36:48.000 --> 00:37:01.300
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.
00:37:01.300 --> 00:37:07.400
So, by use of some cleverness, we were able to skip having to do four simultaneous linear equations,
00:37:07.400 --> 00:37:10.000
to where eventually we are effectively going to bring it down to 3.
00:37:10.000 --> 00:37:16.600
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.
00:37:16.600 --> 00:37:22.100
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.
00:37:22.100 --> 00:37:27.000
We could have solved out those four simultaneous linear equations, because each one will produce things.
00:37:27.000 --> 00:37:33.200
Remember how the first column produced -1 = A + C; each one of these columns will produce an equation.
00:37:33.200 --> 00:37:39.100
And from four simultaneous linear equations, solving for four variables, we will be able to do it; it is just kind of a pain.
00:37:39.100 --> 00:37:43.600
So, we came up with this clever way, and we were able to figure out 2 = B by just happenstance.
00:37:43.600 --> 00:37:51.100
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.
00:37:51.100 --> 00:37:55.300
And we figured out that 2 = B; all right, well, let's work through it now.
00:37:55.300 --> 00:38:01.800
So, we figured out B = 2, and now we have -t³ + 8t² - 6t =
00:38:01.800 --> 00:38:14.100
At³ + Ct³ - 2At² + Bt² - 4Ct² + Dt² + 2At + 4Ct - 4Dt - 4A + 2B + 4D; wow!
00:38:14.100 --> 00:38:20.700
OK, so at this point, let's put columns to columns; so -t³ goes to our At³ + Ct³.
00:38:20.700 --> 00:38:28.900
So, it must be that -1 is equal to A + C.
00:38:28.900 --> 00:38:36.400
We can do this column here; I accidentally cut through that negative sign--hopefully that wasn't too confusing.
00:38:36.400 --> 00:38:53.900
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.
00:38:53.900 --> 00:39:00.500
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.
00:39:00.500 --> 00:39:07.200
And then, finally, the last one of constants: what is our constant? Our constant is 0.
00:39:07.200 --> 00:39:17.300
So, we have 0 = -4A + 2B + 4D; for extra credit, let's just see what this would have been.
00:39:17.300 --> 00:39:22.300
How many t²'s did we have? We had 8, so in our difficult-to-read yellow (I'll make it black,
00:39:22.300 --> 00:39:31.300
just because black is easier to read), 8 = -2A + B - 4C + D.
00:39:31.300 --> 00:39:34.800
We could have worked that one out, but I wanted you to see what it would have been.
00:39:34.800 --> 00:39:39.100
But we will actually end up having enough information, because we have three equations and three unknowns;
00:39:39.100 --> 00:39:41.700
so we have enough with the red, blue, and green things.
00:39:41.700 --> 00:39:48.900
So, at this point, let's figure out A first; so that means we need to solve for everything that isn't A,
00:39:48.900 --> 00:39:52.300
so that we can plug them in, get rid of everything else, and just have A.
00:39:52.300 --> 00:40:04.800
So, on the left side, we have -1 = A + C; so we solve for C, and we get -A - 1 = C.
00:40:04.800 --> 00:40:09.000
OK; on the green one, we plug in what we know for our B.
00:40:09.000 --> 00:40:30.200
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.
00:40:30.200 --> 00:40:39.600
So, we move that stuff over, and we get: add A on both sides; subtract by 1 on both sides; A - 1 = D.
00:40:39.600 --> 00:40:59.700
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.
00:40:59.700 --> 00:41:07.800
-6 = 2(A)...that actually stays around, because it was just A from the beginning; 4...what did we figure out C was?
00:41:07.800 --> 00:41:15.300
C is the same thing as -A - 1, minus 4 times...what did we figure out D is?
00:41:15.300 --> 00:41:29.300
It is the same thing as A - 1; simplify that out: -6 = 2A + 4(-A), so we will make that - 4A, minus 4...
00:41:29.300 --> 00:41:40.400
-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.
00:41:40.400 --> 00:41:53.600
We have -6 = 2A - 8A; -6 = -6A; and thus 1 (dividing both sides by -6) = A.
00:41:53.600 --> 00:41:58.000
With 1 = A in place, we can now easily go back and figure out everything else.
00:41:58.000 --> 00:42:13.000
We have -A - 1, so negative...plug it in, knowing that it is 1...minus 1, equals C; so we have -2 = C; great.
00:42:13.000 --> 00:42:24.800
And also, plug in over here; and we have that 1 - 1 = D; it turns out that D is just 0.
00:42:24.800 --> 00:42:29.400
So, at this point, we have figured out all of the things that we need.
00:42:29.400 --> 00:42:36.800
We have this set up as (t - 2) + (t - 2)² + (t² + 2).
00:42:36.800 --> 00:42:59.000
It was in the format -t³ + 8t² - 6t, all over (t - 2)²(t² + 2).
00:42:59.000 --> 00:43:05.700
OK, so that is the case; we have that as something over (t - 2).
00:43:05.700 --> 00:43:21.100
It was A/(t - 2) + B/(t - 2)² + (Cx + D)/(t² + 2).
00:43:21.100 --> 00:43:28.600
OK, at this point, we can actually plug in our numbers; we have that A is 1, so we have 1/(t - 2).
00:43:28.600 --> 00:43:44.000
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.
00:43:44.000 --> 00:43:53.100
So, we have -2x/(t² + 2); oops, once again, I did it again; I got to thinking in terms of x, as opposed to t.
00:43:53.100 --> 00:43:57.900
But we are using a different variable: -2t/(t² + 2).
00:43:57.900 --> 00:44:02.900
And there we are; we have decomposed it into its partial fractions.
00:44:02.900 --> 00:44:07.000
That is pretty long; it is pretty complicated; but that is pretty much as hard as it gets.
00:44:07.000 --> 00:44:10.700
As long as you break it down, and then you multiply everything out carefully,
00:44:10.700 --> 00:44:16.300
and you are careful with all of your algebra and your arithmetic, you can get it into this form right here,
00:44:16.300 --> 00:44:22.500
where you have this big block compared to these original things, and then you are able to figure out all of these linear equations.
00:44:22.500 --> 00:44:26.900
And you can possibly be clever and figure out some way to figure out B = 2.
00:44:26.900 --> 00:44:32.800
But you also can just work through a bunch of linear equations, solving for one thing at a time in terms of the other,
00:44:32.800 --> 00:44:35.200
plugging them all together, and then solving it out.
00:44:35.200 --> 00:44:37.700
Sometimes it takes a lot of work; sometimes it goes kind of slowly.
00:44:37.700 --> 00:44:40.600
But ultimately, if you just keep working at it, you can get it.
00:44:40.600 --> 00:44:43.000
And it is really, really useful in calculus.
00:44:43.000 --> 00:44:45.600
I know--I know that you aren't going to be using it immediately.
00:44:45.600 --> 00:44:51.500
But honestly, this thing makes a problem that would be, otherwise, totally impossible to do, just really easy.
00:44:51.500 --> 00:44:53.600
So, it is a really handy thing in calculus.
00:44:53.600 --> 00:44:56.000
All right, we will see you at Educator.com later--goodbye!