Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. The goal of partial fractions is to take a fraction and find the pieces that would add up to get the fraction. So if we look at something like five X -, 7 / X ^2 - 3 X +2, the first thing we're going to do is we're going to factor the denominator and X -, 2 and X -, 1. Now if I had some constant over X - 2 and I added that to some constant over X - 1, we could see that the common denominator there would be X2 times X1, which is what it is in the original. And we'd have A and B being single constants or 1° less than the X -, 2 on the bottom. So what we're going to do now is we're going to think about taking that five X - 7 over X - 2 X -1 equaling the a / X - 2 plus the b / X - 1. And we're going to multiply through by the common denominator. When we do that, we end up with five X - 7 equaling a * X - 1 + b * X - 2. So five X - 7 is going to equal AX minus A+ BX -2 B. If we group that right hand side so that we have all the terms with an X together and I'm going to factor out the X and then we have everything that's not with an X together. We can see now that this a + b * X on the right side would have to be equal to the coefficient on the X on the left, and the constant down here would have to be equal to the constant over here. So now we're going to actually have two new equations 5 equal a + b and -7 equal negative a -, 2 B. If we solve this by the elimination method, we're going to add five and -7 to get -2 A and negative a cancel B and -2 B are going to give me negative B. So I know that B is going to be two. If I can find B, I'm going to substitute it into either of those two equations up there to get A as 3. So what this tells me is that in the original, that five X - 7 / X - 2 X -1 is going to really be the same thing as 3 / X - 2 + 2 / X - 1. We could check that if we wanted to. Does 3 * X - 1 + 2 * X - 2, getting them all over that common denominator. If we distribute that out, three X + 2 X is 5 X -3 - 4 is -7. So that is the answer we're looking for. We took one fraction and we made it into two partial fractions that added to give the original. If we look at another example, if we have three things multiplied on the bottom, we're going to have a / X + b / X + 1 + C / X - 5. So if we multiply everything 3 by that common denominator, we're going to end up with four X ^2 - 5 X -15 equaling a * X + 1 * X -, 5 + b times ** -5 plus CXX plus one, multiplying each and every term by the common denominator. So now we're going to do some distribution. We're going to get AX squared -4 AX -5 A+, BX squared -5 BX plus CX squared plus CX. If we regroup things so that we pull all of our X ^2 terms together, we'd have A + B + C * X ^2. We'd have plus -4 A minus 5B plus C all times X, and then we'd have -5 A. We had 123-4567 original terms when we distributed, so we should have 123-4567. So now what we're going to do is we're going to let the coefficient of the X ^2 equal the coefficient of the X ^2. The coefficient of the X equals the coefficient of the X and the constant equals the constant. So 4 equal a + b + C negative 5 equal -4 A minus 5B plus C and -15 equal -5 a. Well, that last one gives us an answer pretty easily. We know that a is going to be 3. So if we know a is 3, we could come up to these other two and we could have 4 equal 3 + B + C, or one would equal b + C We could have -5 equaling -12 -, 5 B plus C So if we add the -12, we're going to get 7 equal -5 B plus C. Now we need to solve for a variable. Let's take this bottom equation and multiply through by a -1. Oops, 1 equal. So we're going to leave the top one the same. We're going to multiply the bottom one by a -1. When I add them, the B's are going to or the C's are going to go away, leaving me 6B equaling -6 or B is -1. And once I get my BI, can substitute it back in to get my A or my C Sorry. So 1 equal -1 plus CC is going to equal 2. So our final answer to this would be 3 / X or A / X + -1 or B over the X + 1 plus our C which was 2 / X - 5. We were trying to find the constants that had to go over those linear terms to make them into three fractions that would add together to give us the original. Now, if they're not linear, oh Nope, this one is linear. So we're going to have a / X, we're going to have B / X + 7, and we're going to have C / X + 7 ^2 because this was a linear to a power. So we actually have to have X + 7 and X + 7 ^2. If you thought about the common denominator between those two, it would be X * X + 7 ^2. So now if we multiply everything through by the common denominator, we're going to get A * X + 7 ^2, we're going to get b * X * 1 of the X plus sevens, and we're going to get C * X. So three X ^2 + 49 is going to equal AX squared plus 14 AX plus 49 A+ BX squared, +7 BX plus CX. Once again, as in the previous examples, I'm going to pull together all the X ^2 terms. So I'm going to get A plus BX squared plus 14A plus 7B plus C all X plus our constant 49 A 123456123456. I usually count to make sure I haven't lost any terms. So the coefficient on the X ^2 has got to equal the coefficient on the X ^2, the coefficient on the X. In this case there isn't any. So 0 has got to equal the coefficient on the X and the constant has got to equal the constant. This bottom one we luck out, we get a equal 1. If we then substitute that one into the top, we get B equal 2 and so now we can put in the 3rd. 114 * a of 1 + 7 * b of 2 plus C has to equal 0. So 14 + 1428 + C = 0 C is going to equal -28 so our final answer here is going to be 1 / X + 2 / X + 7 - 28 / X + 7 ^2. We could always check this by multiplying it out doing our common denominator, and we would get three X ^2 + 49 looking at one that doesn't have all linears. So this is going to be a / X -, 4 just like before. But the next term doesn't factor and it's not a linear, it's a quadratic X ^2 + 5 S in the top. It's going to have to be 1° less, So BX plus C1 less than a quadratic is a linear one. Less than a cubic would be a quadratic. Same exact steps now. So we're going to have the five X ^2 - 9 X plus 19 equaling a * X ^2 + 5 plus BX plus C * X - 4. When we manipulate this out, we're going to get AX squared plus 5A plus BX squared -4 BX plus CX -4 C. Combining our terms again, we're going to have an A+ abx squared plus a negative 4B plus CX plus five A -, 4 C So now we know that the 5 has got to equal A+B, the -9 is going to equal -4 B plus C, and the 19 is going to equal five A -, 4 C I want to choose two of those equations to get one rid of one of the variables. It doesn't matter which two. I think that maybe I will do the B's. If I multiply the top equation through by 4, I'd get 20 equaling 4A plus 4B. If I leave the second one exactly the way it is -9 equal -4 B plus C so we get 11 equaling 4A plus C Now I'm going to combine that with the equation 3 that I hadn't used yet. So if we're going to try to combine 11 equal 4A plus C and the 19 equal five A -, 4 C, let's multiply this top one through by 4. So we would get 44 equaling 16A plus 4C all over Oops, combined with not over 19 equaling five A -, 4 C so the C's cancel, we get 21A equaling 63. So A is going to equal 3 now. It doesn't have anything to do with the fact that it's a whole number. These actually really could easily come out as fractions. If we know A, if we stick it back into the very original 5 equal A+B, we could find B as two. And if I have B as two, if I stick it into one of the others to find our C, we'd get -9 equaling -4 * 2 + C so C is going to equal -1. So our final answer would be 3 / X -, 4 + 2 X -1 / X ^2 + 5. If we do one more example, this is going to be a quadratic squared, so we're going to have ax plus B over X ^2 + 4 plus CX plus d / X ^2 + 4 ^2. So we're going to have X ^2 + 2 X plus 3 equaling AX plus b * X ^2 + 4 plus CX plus D, just multiplying everything through by the common denominator. So X ^2 + 2 X plus 3 is going to equal AX cubed plus BX squared plus 4AX plus 4B plus CX plus D So here we're going to have X ^2 + 2 X plus 3 equaling AX cubed plus BX squared plus 4A plus CX plus 4B plus D. Coefficients have to equal coefficients. There was no X ^3 on the left, so 0 equal a. One coefficient on the left for the X ^2 has to equal the coefficient for the X ^2 on the right. B2 is going to equal 4A plus C and three is going to equal 4B plus D Well, if we know a is 0 and B is 1, then this one's going to equal 2/2 equal 4 * 0 + C or C is going to equal 2. This other 1/3 equal 4 * 1 + d so D is going to equal -1. So our final answer here is going to be 0X, or just 0 + 1 / X ^2 + 4 + 2 X -1 / X ^2 + 4 ^2. You could just write that as 1 / X ^2 + 4 +2 X -1 over the quantity X ^2 + 4 ^2. Thank you and have a wonderful day.