 There are two important extensions of synthetic division that are worth discussing. So let's say I want to divide xq plus 3x squared minus 7x plus 8 by 2x minus 5. And the first thing you should say is, wait a minute! We can only use synthetic division when our divisor is of the form x minus a. Any other use voids our warranty. And that's a good point. Our divisor is not of the form x minus a. So it looks like we're going to have to use the long division algorithm. And while everybody loves doing long division, we might still be able to use synthetic division on this if we remember an important property from the real numbers. A divided by b is the same as a divided by n divided by b divided by n. Since we want a divisor of the form x minus a, we can divide 2x minus 5 by 2. But that will require that we divide both dividend and divisor by 2. So we'll throw in our coefficient of 1 and divide every coefficient by 2. And now since our divisor is of the form x minus something, we can use synthetic division. And now we're go for synthetic division. Drop the leading coefficient. Multiply by 5 halves. We'll need to add to the next column, which means adding two fractions together. So we'll find a common denominator and add. Multiply by 5 halves. Add to the next column. Again, we need to find a common denominator to add. Multiply by 5 halves. Add to the next column. And since this is the last column, we're done and we can read off our answer. One important idea. The last entry here is the remainder. But it's the remainder when we divide by x minus 5 halves. This is not the remainder when we divide by 2x minus 5. However, it's still the numerator of our remainder when we write it in fraction form as long as we use the divisor x minus 5 halves. Which allows us to express our quotient as a polynomial plus a rational expression. Another extension of synthetic division comes from the following useful idea. A polynomial in x is a polynomial that can be expressed as a sum of terms of the form sum coefficient times a power of x. And so this means that something like 3x squared plus 5x plus 7, well that's a polynomial in x. Similarly for something like this, even though the coefficient is horrifying. And importantly, something like this is actually a polynomial in x, because every term is something times a power of x. Now notice that because we have this square root of y here, this is not going to be a polynomial in y, because our power on y is not a whole number. But that does mean that something like this is a polynomial in x, because every term is something times a power of x. And it's also a polynomial in y for the same reason every term is something times a power of y. And that means we can use synthetic division on something like this. So again the important check is that our divisor is of the form x minus something, which it is. For a setup we need to set down the coefficients. So let's set this up as a long division, throwing that coefficient of 1. And our coefficients are everything that is not a power of x. So remember our synthetic division records the coefficients and each column corresponds to some power of x. So we can read our dividend as being 1x cubed plus 3yx squared plus 7yx plus y cubed, which is what we want it to be. And we're ready for synthetic division. Drop the leading coefficient, multiply by y, and add to the next. Multiply by y and add to the next. Multiply by y and add to the next. And that gives us our quotient and remainder. And we can write this. Remember that these correspond to the coefficients of x squared, x, the constant, and the numerator of a rational expression whose denominator is x minus y. Again, we may have to use coefficients of 1, 0, or even negative numbers. So for example, a division like this might begin by rewriting the dividend, so it is a polynomial in x with 1 and 0 coefficients as needed. Our divisor needs to be of the form x minus something, so we'll rewrite that as x minus negative y. And removing our powers of x, we're now ready for synthetic division. So we'll drop our leading coefficient, multiply by negative y and add, lather, rinse. And we're at the end of the table, so we won't repeat this. But we can read off our result, which we should express as a quotient plus a rational expression.