in this video we're going to go over partial fraction decomposition i'm going to show you how to solve many of these problems but before we do that let's talk about what partial fraction decomposition is so what exactly is it well to illustrate it let me give you an example problem let's say if we have 2 over 3x plus 4 over 5y how can we combine these two fractions into one single fraction well we need to get common denominators first so i'm going to multiply the second fraction by 3x over 3x and the first one by 5y over 5y so this will give me 10y over 15xy plus 12x over 15xy now because i have a common denominator i can combine it into a single fraction so i can write it as 10y plus 12x divided by 15xy now what partial fraction decomposition allows me to do is it allows me to take a single fraction like this one and break it down into two smaller fractions so it's the reverse of combining two fractions into a single fraction so you take a single fraction and break it down into multiple smaller fractions it could be two fractions three four it depends on the nature of this particular fraction and so that's the basic idea behind partial fraction decomposition so let's work out an example let's say we have a rational function 7x minus 23 divided by x squared minus 7x plus 10. so how can we take this fraction and break it down into smaller fractions the first thing we need to do is we need to factor this fraction completely we can't factor the numerator however we can factor in this trinomial x squared minus seven x plus ten so what two numbers multiply to ten but add to the middle coefficient negative seven so we know five times two is ten but negative five plus negative two adds up to negative 7. so we have 7x minus 3 divided by x minus 2 and x minus 5. so you need to write the denominator in terms of linear factors and quadratic factors so what exactly is a linear factor linear factors are like x plus two three x minus five four x plus eight x these are linear factors a quadratic factor would look something like this x squared plus 8x plus 3 which is not factorable or it could be x squared plus seven these are quadratic factors now there are some other terms that you need to be familiar with how would you describe this term this is known as a repeated linear factor so x squared is also a repeated linear factor 7x plus 3 squared that's a repeated linear factor now x squared plus 1 squared that's a repeated quadratic factor be familiar with these terms because it's going to affect the way this fraction is decomposed so we have two linear factors and what we need to write is two fractions on this side and each fraction will contain a linear factor now anytime you have a linear factor on the bottom you need to put a constant on top now we're going to choose two different constants a and b our goal is to determine a and b is a equal to three is be equal to four we need to figure that out so what i'm gonna do is i'm gonna multiply both sides of the equation by this denominator that is by x minus two and x minus 5. so if we multiply this fraction by what we have here x minus 2 and x minus 5 will cancel giving us 7x minus 23 on the left side now if i take this fraction and multiply by what i have here the x minus two terms will cancel leaving behind a times x minus five now these two will cancel and i'm gonna be left with b times x minus two now when you get to this part there's two ways in which you could find a and b you could use a system of linear equations or you can plug in x values and determine a and b for a simple problem like this it's best to plug in x values so let's try plugging in x equals five because if we do so this will disappear five minus five is zero and zero times a is zero so then this becomes seven x minus twenty three or seven times five minus twenty three because we need to replace that with five that's equal to a times five minus five plus b times five minus two seven times five is thirty five and five minus five is zero so this disappears and five minus two is three now thirty five minus twenty three is twelve and so that's equal to 3b and if we divide both sides by 3 b is equal to 4. so i'm just going to rewrite that here so now we need to calculate a so this time let's plug in 2 so if x is equal to 2 let's see what's going to happen so we're going to have 7 times 2 minus twenty three and that's equal to a times two minus five plus b times two minus two seven times two is fourteen and two minus five is negative three and two minus two is zero now fourteen minus twenty three that's a negative nine and that's equal to negative three a b times zero is zero so that disappears and now let's divide both sides by negative three so negative nine divided by negative three is three and so that's the value for a now let's go back to our original problem so we had 7x minus 23 divided by x minus two times x minus five and that was equal to a over x minus two plus b over x minus five now we have the value of a a is three and b is four so therefore we could see that this fraction is equal to three over x minus two plus four over x minus five that's the answer now if you ever need to check it simply combine the two fractions and see if they give you what you started with here so let's go ahead and try that i'm gonna multiply this fraction by x minus two over x minus two whatever you do to the top you have to do to the bottom and this one i'm gonna multiply by x minus five divided by x minus five three times x minus five is three x minus fifteen and on the bottom we have x minus five times x minus two which i'm gonna leave it that way and then four times x minus two that's going to be 4x minus 8 divided by x minus 5x minus 2. now if we add 3x and 4x that will give us 7x and then if we add negative 15 and negative 8 that will give us a negative 23. so as you can see the answer that we had at the beginning was correct so it's 3 over x minus 2 plus 4 over x minus 5. that's the answer now for the sake of practice let's try another similar problem so let's say let me rewrite that if we want to decompose 29 minus 3x divided by x squared minus x minus 6. feel free to pause the video go ahead and try that so first we need to factor the denominator by the way notice that the denominator is always one degree higher than the numerator now what two numbers multiplies to negative six and add to the middle coefficient negative one this is going to be negative three and positive two negative three plus two adds up to negative one so we have 29 minus three x divided by x minus 3 times x plus 2. so let's set this equal to a over x minus 3 plus b over x plus 2 since we have two linear factors now just like before we're gonna multiply both sides of this equation by this denominator so that is by x minus three times x plus two so these two will cancel and on the left side we're going to get twenty nine minus three x now if we take a over x minus 3 and multiply it by this we can see that x minus 3 will cancel and then we're going to get a times x plus two and then if we take this fraction multiply by x minus three times x plus two the x plus two factors will cancel leaving behind b times x minus 3. actually let's keep that there for now so at this point let's plug in x equals three this is going to be 29 minus three times three and that's equal to a times three plus two plus b times three minus three three minus three is nine and three plus two is five three minus three is zero so this disappears 29 minus nine is twenty and that's equal to five a so if we divide both sides by 5 we can see that a is equal to 4. i'm going to put that in the side over here now let's plug in negative 2 so that this term becomes 0. so we're going to have 29 minus 3 times negative 2 and that's equal to a negative 2 plus 2 plus b times negative two minus three now negative three times negative two is positive six negative two plus two is zero and negative two minus 3 is negative 5. 29 plus 6 is 35 and so that's equal to negative 5b so b is going to be 35 divided by negative 5 which is negative 7. so now that we have a and b we can now write the final answer so let's plug it in to this expression so 29 over i mean 29 minus 3x over x squared minus x minus 6 that's equal to 4 divided by x minus 3 and then b is negative 7. so instead of writing plus negative 7 i could simply write minus 7 because a positive number times a negative number is still a negative result so it's going to be minus 7 over x plus 2. and this is the answer you