 Let us compute the partial fraction decomposition for 9x minus 15 over x cubed minus one So this is the proper fraction We have a linear polynomial on top a cubic on the bottom if I tried to factor the denominator Which I need to do I notice that x cubed minus one is a difference of cube So I get x minus one and x squared plus x plus one Like so and this is sits below x 9x minus 15 For which if you try to factor x squared plus x plus one It's actually irreducible the discriminant of that thing is going to be a negative three Which that really can't factor that with real numbers anymore. That's okay. We'll just adjust our We're gonna adjust our template to accommodate for the fact that way this irreducible quadratic Right in which case the your template would look like a over x minus one And then your second denominator would be x squared plus x plus one like so But what's the numerator supposed to be because after all these are supposed to be proper fractions, right? If my denominator is quadratic then your numerator could be well It could be linear actually so we have to have something like bx plus c like so In which case my strategy is going to still be let's clear the denominators times both sides by the LCD So if you cleared the denominators as we multiply both sides by x minus one and x squared plus x plus one We do that to the left-hand side as well times both sides by x cubed minus one. They will cancel out the numbers on the left Leaving you just the numerator 9x minus 15 and then on the right-hand side You're gonna get a times x squared plus x plus one and then you're gonna add that to bx plus c Times x minus one like so for which then we proceed to annihilate if we check x equals one, right? That'll annihilate b and c at the same time. So we're gonna plug in one here We're gonna get 9 minus 15 on the left-hand side, which is equal to in that case negative 6 On the right-hand side, you're gonna get 1 plus 1 plus 1 for the coefficient of a so you get 3a So dividing both sides by 3 We end up with a equally negative 2 That's the first observation. What do you do about the b and the c right? What do you plug into x squared plus x plus 1 to cancel it out? Well, those are actually to be non real complex numbers So if we want to do some arithmetic with complex numbers, we could do that or we're gonna take it the following approach Let's plug in let's actually plug in negative 2 for a right We know it is equal to negative 2 now from the calculation. We just did so if we plug in negative 2 and We do the same thing right here. Let me just get rid of all of these parentheses and things. I'm just gonna rewrite it So once we discover that a is negative 2 we can plug that in there So we get negative 2 right there and that's supposed to still equal 9x minus 15 over x cube minus 1 So what do we do about the b or the c right? Well, notice b is attached to x if I were to plug in x equals 0 That would annihilate the b it doesn't annihilate the a but hey a is negative 1 or negative 2 I can live with that in which case the left-hand side will be a negative 15 We plugged in x equals 0 then we have a negative 2 times 0 plus 0 plus 1 And then we're going to get 0 plus c times negative 1 All right, simplifying what we can here we get negative 15 is equal to negative 2 minus c So we're gonna add 2 to both sides. We get negative 13 equals negative c. So c equals 13 And so we're then gonna make that substitution in above Okay, so since c turned out to be 13 I'm gonna erase that and plug in 13 So how do I figure out b at this moment? Well, we know what C is c is a 13 We plugged in and got a so we plugged in 0 we plugged in 1 we got a and c here to figure out b Let's just plug in something else We should plug in something probably pretty easy to do arithmetic with maybe like negative 1 or 2 Those are both fairly safe examples here. I'm gonna do x equals 2 again I want the arithmetic to be simple here x equals 2 on the left-hand side You're gonna get 18 minus 15 which turns out to be 3 on the right-hand side. You're gonna get negative 2 times 2 squared is 4 plus 2 plus 1 We're then gonna get 2b plus 13 times 2 minus 1 which is a 1 and So what do we have here? We have like I said 3 we're gonna get negative 2 times 4 plus 2 plus 1 is a 7 Then we have 2b plus 13 If I subtract 13 from both sides, you're gonna get negative 10 is equal to negative 14 plus 2b If you add 14 of both sides, we get 2b is equal to 4 and therefore b is equal to 2 So it took a little bit more detective work to figure out the coefficients this time But if I come up here and then fix my coefficient b turned out to be 2 We now have our partial fraction decomposition So I have to say that Irreducible quadratics do make the process a little bit more challenging because we can't we can't quite as easily annihilate terms But this process of annihilation will still work and we can find the partial fraction decomposition for these rational expressions