The A Star way to factorise cubics

If you truly want to master factor theorem, then you need to be able to do this. Instead of doing algebraic division and wasting your time, you should be able to factorise and solve these types of things really, really quickly. And I'm going to show you how to do it. You might see it online called factorisation by inspection, but no one ever explains how to do it. That's what I'm going to do. So there are three examples. This is really important as well because it comes up a lot on university entrance exams. For example, the TMUA as an example and the ESAT, where timing is a really big deal and you don't want to be doing algebraic division 50 times over. So anyways, let's get started with one of these questions. So it's given that x plus 1 is a factor of this, p is a positive constant, find the value of p. Well, that part's not too bad. You just sub in f is minus 1, and then you just rearrange, right? That's completely fine, and you eventually get that p is equal to 9. So now I'm going to show you the actual part I want to talk about, which is how on earth do you factorise this without having to do the long process of algebraic division? And we're actually going to do it also with, for example, quartics and quintics as well, just to show you that this works for a lot of different things. Well if I know that this is a factor that means I can divide this expression so this whole thing here by x plus 1 and my result will be x plus 1 and then some stuff in the bracket. That stuff in the bracket should be a quartic sorry a quadratic because think of it if I'm dividing an x cubed by an x I'm left with x squared. If there was an x squared inside this factor then I'd get a linear expression on the inside okay. So essentially, you just divide the first power here by the first power here, so we get a quadratic. So I know it's going to have x squared, it's going to have x, and it's going to have some number on the end. But the question is, what is it going to look like? Well, first things first, x times x squared would give me x cubed, but we want 2x cubed. The only way for me to get that is to shove a 2 in front of the x squared. It's the only way it works. Okay, cool, so x times 2x squared is 2x cubed, perfect. What else can we do? Well, the only way I can get a constant of minus 18 is by doing 1 times the number on the end here. So 1 times what gives me minus 18? Well, it's minus 18. Okay, I'm going to go through one last example showing you that factorization by inspection. So again, they tell me x minus 6 is a factor of this, so I'm going to write this as x minus 6. And again, what's the remainder? Well, 2x cubed divided by x gives me x squared, so I'm going to write a quadratic. Remember, if they give you a factor as x squared or something, you'd have to minus the power by 2 and so on. which I've seen them do before. And we don't know anything to do with x. And we don't know anything to do with the number on the end. So I'm literally just going to leave them blank. And in fact, I'm even going to remove the plus signs because they could be negative. And all we're going to do really simply is, in my opinion, start with the number in the front. So I know that x times something x squared gives me 2x cubed. So that number has to be a 2, right? Cool. That's progress already. What else do I know? Well, I know that minus 6 times the number on the end should give me 18 so minus 6 times blank gives me 18 that blank has to be minus 3 there you go already know two of the three terms now how did I work out the number in the middle again well all we can do is look at this and get minus 9x what two ways do I get that well I have minus 6 times whatever this is so why don't we call it a is just a placeholder so minus 6 a x and then we also get it from doing x times this so minus 3x should give me minus 9x so probably the easiest way is to move this over and cancel out the x's in the same step we get minus 6a equals minus 6 yeah so we get a is equal to 1 so literally my quadratic there is just that 2x squared plus x minus 3 and then factorizing this is super easy times to minus 6 adding to 1 so it's going to be 3 and minus 2 x minus 6 so you're to get 2x plus 3 which is completely fine and then you're going to get 2x minus 2 but you can simplify that so just x minus 2 that is a lot faster than factor than doing the algebraic division the other reason why i like this method more is because when you do algebraic division it is so easy to make mistakes even myself if i'm doing something like x to the 4 divided by something like this it's really easy to mess up with the signs this method is foolproof because what can I do? I can very quickly expand this and see if I'm right because that's how I'm doing it in the first place. So again that's a nice little trick you can use to help you factorize things a lot faster.