 Two of the most important results from precalculus are known as the root and remainder theorems. In order to understand these, we'll need to take another look at division. Consider a division of a whole number that gives a quotient and remainder, like 137 divided by 7 equals 19 with remainder 4. One way we can look at this division is we started with 137 and we subtracted 7, 19 times, at the end of which we had 4. Another way to look at this runs backwards. If we start at 4 and add 7, 19 times, we'll get 137, which says that 137 is 197s plus 4. This is generally true for any division and is based on our definition. Suppose a divided by b is q with remainder r, then a is equal to bq plus r. And what this means is that any time we have a quotient and remainder, we can rewrite it as a product and sum. For example, suppose you're out walking and out of the sky, 140 divided by 11 equals 12 with remainder 8 falls and hits you on the head. Because you know the definition of a division with quotient and remainder, you can immediately rewrite this as a sum and product. So comparing our statement to our definition we have, and we can say that 140 is 11 times 12 plus 8. Well, math ever generalizes. This definition, which we used initially with whole numbers, extends to polynomial quotients as well. And this is useful in the following way. If capital X is a degree n polynomial and a is any number, we can divide capital X by x minus a to get an n minus 1 degree polynomial and a constant remainder. We can then rewrite our original polynomial capital X as a sum and product. So for example, if you know the result of a polynomial quotient, we can rewrite this in product and sum form. Don't worry, this result is not going to fall out of the sky and hit you on the head. So let's pull in our definition of quotient and remainder. So this says our original dividend is equal to the product of the divisor x minus 2 with the quotient x squared minus 10 x minus 14 plus the remainder negative 23. How about trying to evaluate this when x equals 2? So remember, equals means replaceable, so every time I see x I'm going to replace it with 2. Again, equals means replaceable, so I can either find the left-hand side or I can find the right-hand side. If we take a look at the right-hand side, the first factor, 2 minus 2, is going to be 0, which means that this entire first term will multiply to 0, and so my right-hand side will be 0 plus negative 23 or just negative 23. And since equals means replaceable, it means that if I went through the trouble of calculating the left-hand side, I would end up with negative 23. But I was able to find it much more easily. And this suggests what's known as the remainder theorem. Let capital X be any polynomial. To find capital X when x is equal to a, find the remainder when capital X is divided by x minus a. For example, let's find the value of x cubed minus 37x squared plus 24x plus 11 when x equals 2. Well, since the remainder theorem is our shiny new toy, let's use it. So if we want to find the value of this polynomial when x equals 2, the remainder theorem says find the remainder when you divide. Well, there's just one problem. We have to do the division. So unless you can divide quickly and efficiently, don't use the remainder theorem. Fortunately, we do have a quick and easy method of doing a division of this type. We can use synthetic division. So we'll divide x cubed minus 37x squared plus 24x plus 11 by x minus 2. And so our polynomial is negative 81 when x equals 2.