 Now, if I wanted to divide two polynomials, I generally have to use some sort of long division algorithm. But in certain cases, we can use what's called synthetic division. Now, before we look at synthetic division, I've been told that some people would like to know what I look like. Now, those of you who are actually taking a class from me know what I look like. But for everybody else, I dug through my files and I found the following picture showing me giving a lecture on how to find the volume of a sphere. Okay, so on to the mathematics. So, the reason that synthetic division is a very, very, very limited algorithm is that it only works when I'm dividing a polynomial by a divisor in the form x minus a. And it's very important to understand this. So, for example, let's say I want to do this division. My divisor is not x minus a, so I can't use synthetic division. Or if I have something like this, again, my divisor is not x minus a, so I can't use synthetic division. Finally, let's say this quotient. My divisor is x minus a, so this is something where I can use synthetic division. So, how does this work? We'll take a look in class to see why this works, but the procedure looks something like this. So, the first thing I want to do is I only want to record the coefficients of the terms of the dividend. So, here's my dividend. It's the thing that I am dividing. And so, it's helpful to label our columns by the degree of the term. So, here's our cube, x squared, our x, and our constant term. And I want to set down my coefficients. So, my x cubed term, the coefficient of that is 1. My x squared term, the coefficient is minus 3. My x term, my coefficient is plus 5. My constant term, that's going to be 7. So, there's my setup. And then the other thing I want to do is I want to find the a value of the divisor x minus a. So, my divisor x minus 4, that says that a is the same thing as 4. So, I'll set that down over here in the space that I have reserved for it. Okay, so now the first thing I'm going to do is I'm going to drop that leading coefficient straight down. This one goes straight down, and then I'm going to multiply by this value here. So, 4 times 1 is 4. I'll set that down to the next column, and then I'll add. So, negative 3 and 4, I add that, get 1, and then I'll repeat this process. So, I'm going to go 4 times 1 gives me 4, and I'll set that down to this column, and then I'll add 4 times 9 is 36, and then I'll add, and now I've reached the last column. Now, to read the quotient, the thing to recognize here is that my quotient is going to be a polynomial with a degree 1 less than that of the original dividend. So, this x cubed column becomes an x squared column. This x squared column becomes an x column. This x column becomes a constant. And this last column, this thing that used to be our constant column, that works out to be what our remainder is. So, here my quotient is going to look something like that. Now, I do want to answer the question in the same dialect it was given in. So, I have a polynomial divided by a polynomial. I would like to answer in terms of a polynomial. And so, at this point, I could pick off the coefficients 1x squared, 1x plus 9, with the remainder of 43. So, I can say that my quotient, this divided by that, 1x squared plus x plus 9, and my remainder is going to be 43. Alternatively, remember that any time we have a remainder, I can express that as a fraction, the remainder over the divisor. So, this quotient is also x squared plus x plus 9 plus 43 over x minus 4. And so, there's an alternative way of expressing that quotient. Now, the most important thing about synthetic division is that we can't divide anything other than x minus a. So, something like this, this is not a synthetic division type problem. We have to do something else. And this, x cubed minus 4, x minus 5 divided by x plus 3. Well, the divisor is not in the form x minus a, so we can't use synthetic division. Or can we? Well, we might use a few of the rules of operating with the real numbers. x plus 3, any time I subtract a negative, it's the same as adding. So, x plus 3 is the same as x minus negative 3. And so, even though this divisor doesn't look like it's something we can use, it is a thing we can do using synthetic division as long as we change it to x minus a negative 3. And here's the important thing, we have to make sure that our divisor hasn't actually changed. The only difference is it is in a different form, but it's still the same thing. So, I can now go on to setting up my synthetic division table. So, again, it's convenient to set down the columns indicating which coefficients we're working with. So, I have x cubed, so I have my x cubed, my x squared, my x, my constant, and I want to write down the coefficients of my dividend. So, for x cubed, my x cubed coefficient is 1, for x squared, my x squared coefficient is, is, is... There's no x squared coefficient. So, that says my x squared coefficient is going to be 0. My x coefficient, number in front of x, that's minus 4, and my constant minus 5. So, here's an important thing every now and then we'll have a column that has a 0 in it because we don't have that coefficient in our dividend. And then the next thing is we need the a value of our divisor x minus a. So, our divisor x plus 3, that's x minus negative 3, so my a value is going to be minus 3. And now, I'm ready to apply my synthetic division algorithm. I'll drop the leading number, I'll multiply negative 3 by 1 is negative 3, I'll add, negative 3 times negative 3 is 9, I'll add, I'll multiply negative 3 by 5 is negative 15 and add. And I've reached the last column, so now I am done with the synthetic division and I can write my answer. And again, the terms I get here are the coefficients of a polynomial with a degree 1 less than what I started with and the last term is going to be my remainder. So, this is an x squared, this is x, the x column becomes a constant column, that last column becomes my remainder and my quotient is going to be this divided by this gives you polynomial 1x squared minus 3x plus 5 remainder minus 20. So, there's x squared minus 3x plus 5 remainder minus 20 or I can also again, anytime I have a remainder, I can always express that as a fraction remainder over divisor. So, alternatively my quotient x with minus 3x plus 5 plus negative 20 over x plus 3.