 So here's a little confession. A long time ago I thought synthetic division was kind of a pointless algorithm. You can only ever use it to divide by x minus a, and what good is an algorithm that you can only use for one thing? Well, it turns out there is no use for an algorithm that you can only use for one thing. However, dividing by x minus a is important enough that if you have synthetic division, there are other things that we can do with it. So let's consider this problem, quotients and remainders, so let's remember that if I do a division and find a remainder, I can rewrite this as a quotient equal to the quotient 2 plus the remainder over the divisor. So 13 divided by 5 equals 2 remainder 3. I can rewrite that as 13 fifths is 2 plus 3 over 5. So similarly, algebra is generalized arithmetic, so anything we do in arithmetic, turning a quotient into a statement about fractions, we can also do in algebra. So if I do this division and I find I get quotient x squared plus x plus 9 remainder 43, I can rewrite this as quotient plus remainder as a fraction 43 over x minus 4. Now, I'll do a little bit more algebra here. If I multiply both sides by x minus 4, over on the left hand side, I am dividing by x minus 4, so that multiplication by x minus 4 cancels out. I just have the numerator. Over on the right hand side, I have x minus 4 times each of the terms here. So I think that x minus 4 times my quotient plus x minus 4 times this is going to leave me with 43. Well, okay, who cares? Doesn't seem particularly exciting or relevant that we have this particular algebraic fact, but let's dig in a little deeper. So suppose I am walking down the street one day, and a sign falls out of the sky and hits me on the head. And the sign says, x cubed minus 3x squared plus 5x plus 7 is equal to x minus 4 times x squared plus 9 plus 43. So suppose this comes out of the blue and hits me and I say, ah, I know this has a fact. And then following that sign, a piece of paper flutters down and says, find the value of this polynomial at x equals 4. Well, there's two different things we could do. Now, one of the things we can do, what it means to find the value at x equals 4 is find out what this expression is if I let x be 4. So the obvious thing to do is to let x be 4 and evaluate the expression for cubed minus 3 times 4 squared plus 5 times 4 plus 7. I can evaluate this expression. On the other hand, I did just get hit with this sign. So maybe I can do something with that. Maybe the sign is a sign. So, well, here's something. Because these are equal, whatever I have, whenever I see this, I can replace it with this. They are equal and identical. And if I'm going to drop x equals 4 into this, I can also drop x equals 4 into that. So if I drop x equals 4 into it, I get 4 minus 4 plus 4 squared plus 4 plus 9 plus 43. And a little analysis goes a long way before we actually try to evaluate this. The thing we might notice here is 4 minus 4 is 0. So I don't actually care what the value of this is. Whatever this is doesn't make a difference because it's going to be 0 times who cares what. This first term here drops out and the other thing I've left over is the 43. And so what that tells me is that if x is equal to 4, this expression is equal to just 43. And this is an example of what's called the remainder theorem. Suppose I take a polynomial divided by x minus a and I get another polynomial and I have my remainder. Then when I evaluate that polynomial at a, what I get is just the remainder all by itself. So for example, let's say I want to evaluate 3x cubed plus 5x squared minus 4 at x equal to 4 thirds. Well, what I could do is I could drop 4 thirds into here and I'll have an expression involving a fraction and another expression involving a fraction and then I'll subtract 4. So I'll have to do a whole bunch of fraction arithmetic to evaluate this at x equals 4 thirds. On the other hand, I can use the remainder theorem. And what that says is I want to find the remainder when this is divided by x minus 4 thirds. So this is the perfect type of problem for synthetic division. So I'll set up my synthetic division grid. So my divisor x minus 4 thirds, so my a value is going to be 4 thirds. Now, that's the same as what I'm evaluating at. My coefficients, well, this is an x cubed polynomial. So I have my x cubed, x squared, x and constant column. My coefficients of x cubed is 3x squared. Coefficient is 5x. Coefficient is nothing here. So that's going to be 0. And my constant term is going to be negative 4. So I'll set up my synthetic division grid and I'll perform that division. So following the synthetic division algorithm, I will drop that leading coefficient. I'll multiply and record. So 4 thirds times 3 is 4. I'll add. I'll multiply. 4 thirds times 9 is 12. I'll add. I'll multiply. And then add. And remember the remainder is going to be the last term here. So again, my remainder theorem says that if I want to evaluate this at x equals 4 thirds, I'm going to divide by x minus 4 thirds, and the value is going to be what my remainder is. So my remainder is the last term. So at x equals 4 thirds, this expression is going to be equal to whatever the remainder is, is going to be equal to 12.