This section talks about the remainder theorem and how to use it to solve mathematical problems.

**The Basics**

But before we dive into the remainder theorem, lets first consider a few basic concepts regarding polynomial operations:

In the equation:

**f(x) **is the divided

**d(x) **is the divisor

**q(x) **is the quotient

**r(x) **is the remainder

let’s now take an example:

The **f(x)** in this example is **2x ^{2}- 5x- 1**

The **d(x)** in this example is **x- 3**

After solving the equation:

**q(x) equals 2x+ 1** and **r(x) equals 2**

Let’s try another example:

Can you Identify **f(x)**,**d(x), q(x) **and **r(x)** in this equation?

**Putting It All Together**

Going back to our initial formula of **f(x)****= d(x).q(x)+ r(x)**

We can write:

**2x ^{2}- 5x- 1 **

**= (x-3)(2x+1)+ 2**

**X ^{3}-7x-6= (x-4)(x^{2}+ 4x+ 9)+ 30**

**The Remainder Theorem**

*If a polynomial **f(x) **is divided by another polynomial **(x- c) **the **remainder is always equal to f(c)*

What this means is that **(x-c) **is now the divisor **d(x)**while **f(x) **remains to be the divided

I.e. **f(x)****= (x-c).q(x)+ r(x)**

For this to be true **(x-c) **must be of degree 1, while **r(x) **is of degree 0. In simpler term **r(x) **is just a constant **R**

Now let’s assume that the value of **x **is the same as that of **c**.

If **x= c **then

**f(c) = (c−c)·q(c) + R**

**f(c) = (0)·q(c) + R**

**f(c) = 0 + R**

Therefore:

**f(c) = R** Where R is the remainder and this is the Reminder Theorem.

**Using the Remainder Theorem to Work Out Remainders in Polynomial Operations**

Now let’s use the remainder theorem to work out the following problems

**Question 1**: What is the remainder of:

Note that **c** in this equation is **3**

Therefore, to calculate the remainder [**R** or **f(c)**]

we simply calculate f(3) as follows. (Putting x=3 in the equation)

2(3)^{2}− 5(3) −1

(2x9) – (5x3) −1

18−15−1

=2.

The remainder is 2

**Question 2**: What is the remainder in the equation:

4^{3}− 7(4)− 6

64− (7x4)− 6

64−24−6

= 30

The remainder is 30

**Example 2**

The remainder theorem shows that the value of **f(x)** at **x=c** is the same as the remainder **R** and we’ve already established that **R= f(c)**

Therefore, you might find a question asking you to find the value of **f(x)** as shown below:

**Question 3**: Using the Remainder Theorem work out **f(x)** = 2x^{2}- 5x- 1 at x = 3.

The solution to this problem can be found by working out f(c) as described earlier

2(3)^{2}− 5(3)− 1

(2x9) − (5x3) − 1

18−15−1

= 2

The answer can be given as **f(3) = 2**