Synthetic division of polynomials
What I want to do now is simplify the exact same expression but do it with traditional algebraic long division. And hopefully we'll see why synthetic division actually gives us the exact same result. We'll be able to see the connections between synthetic division and algebraic long division. So let's get started. So if we're doing algebraic long division and set it up right over here, the first thing we want to think about is how many times does the highest degree term here, which is our x, go into the highest degree term here, which is our 3x to the third power. Well x goes into 3x to the third power 3x squared times. So we'll write it in the x squared place, 3x squared times. Now you might already see your parallel. When we did the synthetic division, we dropped this 3 straight down, and this 3 represented 3x squared. So this 3 and this 3x squared are really representing the same thing. But you might be saying, well here we have to do some thinking. We had to say x goes into 3x to the third 3x squared times. Here, we just mindlessly dropped this 3 straight down. How did that work? The reason why we were able to mindlessly drop this 3 straight down is because we assumed, to do this basic type of synthetic division, that we had just an x right over here. We didn't have a 3x. We didn't have a 4x. We didn't have an x squared. We just had an x. And if you divide an x into whatever this highest degree term is, your coefficient is going to be the exact same thing. It's just going to be one degree lower. So you go from a 3x to the third to 3x squared-- so the exact same coefficient. And now, in our little synthetic division right over here, that right over there is the x squared term. So you went from 3x to the third to 3x squared. We essentially divided it by x. And we could blindly do it because we knew, we assumed, that we were dealing with just a 1x to the first right over here. But let's keep on going and see the parallels and see why we're essentially doing the exact same thing. Now let's take this 3x squared and multiply that times x plus 4. So 3x squared times x-- I'll do it in that white color-- it's going to be 3x to the third. And 3x squared times 4 is going to be 12x squared. And now we'll want to subtract this. So now we subtract. We subtract this out. These guys cancel out, and you have 4x squared minus 12x squared. And so that will give you negative 8x squared. So once again, you're probably seeing some parallels. You had the 4x squared over here. You have the 4x squared over there. We just wrote the coefficient, but that's what it represented. 4x squared, we wrote the 4 there. Then we essentially subtracted 12x squared. And the way we got that 12, we multiplied 3 times 4, and then we subtracted. Here we're multiplying 3 times negative 4. We're essentially multiplying 3 times 4 and then subtracting. That's why we put that negative there, so we don't have to keep remembering to subtract this row. So we could just keep adding them. But that's essentially what we did. We multiplied 3 times this 4. And now we subtract it. We get that negative 12x squared. And then we subtracted, and you got negative 8x squared. And you might say, oh, is this the same negative 8 as this right over here? Not quite yet, because over here, this negative 8 literally represents negative 8x. This is actually part of our simplification. When we divide this into that, we got 3x squared minus 8x plus 30. So over here in the algebraic long division, we then say, how many times does x plus 4 go into negative 8x squared? Well x goes into negative 8x squared negative 8x times. So this is the key right over here. x goes into it negative 8x times. And once again, the reason why we could just put this negative 8 here is we know that we are dividing just by a 1x. So you're going to have the exact same coefficient, just one degree lower. So this right over here is our x term, and you see it right over there, just like that. So a lot of the simplification just comes from the idea that we are assuming with the synthetic division that this is a 1x. But let's just keep going. So you have a negative 8x times this business right over here gives you negative 8x times x is negative 8x squared. And then you have negative 8x times 4, which is negative 32x. And we can bring down all of this business right over here, just so it becomes a little bit simpler. So you have a negative 2x. And then over here, you have a minus 1. And once again, when you're doing traditional algebraic long division, you're going to subtract this from that up there. So if we're going to subtract, that's like adding the negative. These characters cancel out. You have a negative 2x plus 32x. That gives us a positive 30x. And then we can bring down that negative 1 if we want. I'll do that yellow color, actually. We'll bring down that negative 1. So this 30 has the same coefficient here. But this 30 should be up here. This is going to be part of our final answer. And to get that, once again, it all comes from the fact that we know that we had an x here when we did the synthetic division. 30x divided by x is just going to be 30. That 30 and this 30 is the exact same thing. And then we multiply. 30 times x is 30x. Actually, let me write the 30x in that white color because that's the convention I've been using-- 30x. 30 times 4 is 120. And then we are going to subtract this from that. So we get negative 1 minus 120 is negative 121, which, right over there, is our remainder-- which is exactly what we got over there. So hopefully you see the connection. Because we are assuming that we are dividing by x plus or minus something, we were able to make some simplifying assumptions. Whenever you divide this by an x, you know it's going to have the same coefficient, just one degree lower. And we kept doing that. And so it allowed us to do it a little bit simpler, a little bit faster, and using less space.