Remainder Theorem and Factor Theorem

We started Chapter 2 with long division. Easy, right? Well this time we learned dividing a polynomial function with a binomial. In the following example, P(x) is divided by the factor x-4.

The quotient for the example above is  x² + 4x + 9 with a remainder of 30. Thus, x³ – 7x – 6  =   (x – 4) (x² + 4x + 9)  +  30. When dividing, just remember to do so carefully to avoid mistakes. 

The Remainder Theorem is an easier way to find the remainder. Instead of using the long division method, take x=b from the factor x-b and substitute it in the polynomial function. From the example above, b=4 so we solve P(4) to find the remainder. 

P(4) = (4 – 4)((4)² + 4(4) + 9) + 30 

                                                              = (0)(16 + 16 + 9) + 30       

                                                              = 0 + 30       

                                                              = 30

Therefore, the remainder theorem:

When you divide a polynomial P(x) by x-b the remainder R will be P(b).

Now, if the remainder is 0, then x-b is a factor of the polynomial. Just as with the Remainder Theorem, the point here is not to do the long division of a given polynomial by a given factor. The same method is used as the Remainder Theorem simply by substituting b to find P(b). The Factor Theorem is just a result of the Remainder Theorem. It's purpose is to check for a zero remainder to determine the factors of the polynomial. 

And so we have: 

When P(b)=0 then x-b is a factor of the polynomial P(x).

Knowing that x-b is a factor of P(x) is useful as it tells you that b is an x-intercept of the function.

If you are still confused, the video below explains the difference between the Remainder Theorem and the Factor Theorem.

In addition to few examples of both the Remainder Theorem and Factor Theorem, this websites provides an exercise for you to work on. Yes, answers are given. I hope this post has helped you in a way. Thanks for reading. :)