 Now, in general, when we find the quotient of two polynomials, we have to use long division. But in certain very specialized cases, we can use what's known as synthetic division. Synthetic division is a shorthand form of long division. And the most important thing to remember about synthetic division is that synthetic division can only be used on quotients where the divisor is of the form x minus a. Now we can give a verbal description of how to perform synthetic division, which is about as useful as a verbal description of how to play the flute. But a good strategy for learning something and for checking your understanding of the topic is writing about it. So I think I understand synthetic division, and here's how I'd describe it. We can divide by x minus a using the synthetic division algorithm. First, we'll record the coefficients of the dividend as a sum using negative, zero, and one coefficients as necessary. We'll also set down a. That's the thing we're subtracting from x to form our divisor. Drop the leading coefficient. Multiply by a and add to the next column. And we'll repeat this process until we're done. At this point, we'll read off the quotient. The remainder is the last term, and the coefficients of the quotient are the other terms in order. Again, until you actually do synthetic division, this makes about as much sense as instructions for playing the flute. So let's do some synthetic divisions. So let's see if we can do this division. So we'll check the fine print. We can only use synthetic division if our divisor is of the form x minus a. And it is. It's worth remembering that synthetic division is a form of long division. So if we imagine setting up our long division algorithm. Now for the synthetic division algorithm itself, we only need to record our coefficients. But it's helpful to indicate at the top of each column what term they correspond to. So this five corresponds to the x squared term, this three to the x term, and this minus seven to the constant term. Meanwhile, this two indicates that our divisor is x minus two. Now we'll drop the leading coefficient straight down. We'll multiply this by our a value and get 10, which we'll add to the next column, which gives us 13. We'll multiply this by two to get 26. Add this to the last column to get 19. And since that's the last column, we're done. And we can read off the result. So the last term is our remainder, 19, and the other two terms give us our quotient in descending order. That means this term must be a constant, and this term must be the coefficient of the x term. And so we can express our quotient as, and remember we can always express our quotient and remainder by expressing the remainder as a fraction over the original divisor x minus a. Or let's take a look at a much more horrifying division. It's helpful to start by writing this as a long division problem, and we have to check the fine print. Synthetic division only works when our divisor is of the form x minus a. But we're dividing by x plus two. So what can we do? Here we can rely on a rule of working with real numbers. x plus two is the same as x minus negative two. And now we can treat this as a synthetic division problem. So we'll put in our coefficients of one, and we'll drop all the surplus baggage. Again, it's helpful to remember that our first column corresponds to the x cubed terms and so on. Now we're going to be multiplying things by negative two, so we'll set that aside to get ready. We'll drop our leading term. Multiply by negative two, then add it to the next column. We'll multiply the sum by negative two, then add the result to the next column. Again, we'll multiply the sum by negative two and add it to the next column. And since this is the last column, we're done. And we can read out the quotient. Remember, the last term is the remainder, and the other terms correspond to the coefficients of the quotient beginning with a degree one less than the original dividend. So these terms are 1x squared, 9x, and 33 with a remainder minus 120. And again, we can also write the remainder as a fraction over the original divisor. And let's make this look at least a little bit neater. We don't typically write coefficients of one, and plus a negative and minus a negative can also be cleaned up. One of the important things to keep in mind about synthetic division is we need to record all coefficients using negative, zero, and one coefficients as necessary. So here, our dividend has an x to the third and an x term, but no x squared term. We say that it's a depressed polynomial. Possibly it's depressed because it's missing something. So again, synthetic division is essentially long division written short. So we can set up our long division. And we'll rewrite our dividend using negative, zero, and one coefficients as necessary. Now, we only need the coefficients of the subtrahend of the divisor, so we'll get rid of everything else. But again, it's useful to remember that our first column represents our x cubed terms, the second the x squared terms, and so on. So we'll need to multiply things by 3. So we'll get that ready. We'll drop our leading coefficient, then multiply this by 3 to get, which we'll add to the next column. We'll multiply our sum by 3, then add this to the next column, multiplying our sum by 3 and adding to the next column. And since this is the last column, we're done. And we can write our result. The last term is the remainder. And the other terms are the coefficients of our quotient constant x term x squared term. Alternatively, we can write the remainder as a fraction over x minus 3.