Remainder Theorem

In this tutorial, we will learn about the polynomial remainder theorem. But before we can do that, there are a couple of things we need to address first.

Polynomials and Their Notation

Let’s begin with an example of a polynomial and its notation.

So here we have a trinomial – a polynomial with three terms – and we denote this polynomial with the letter $\hspace{0.2em} P \hspace{0.2em}$. We generally denote polynomials using a letter from the English alphabet, often $\hspace{0.2em} P \hspace{0.2em}$ or $\hspace{0.2em} Q \hspace{0.2em}$.

And the $\hspace{0.2em} x \hspace{0.2em}$ inside parentheses indicates that the variable in the polynomial is $\hspace{0.2em} x \hspace{0.2em}$. So $\hspace{0.2em} P \hspace{0.2em}$ is a polynomial in $\hspace{0.2em} x \hspace{0.2em}$.

Value of a Polynomial

To talk about the value of the polynomial if $\hspace{0.2em} x \hspace{0.2em}$ is replaced with some constant, we write that constant inside the parentheses (in place of $\hspace{0.2em} x \hspace{0.2em}$).

So, $\hspace{0.2em} P(2) \hspace{0.2em}$ refers to the value of the polynomial $\hspace{0.2em} P \hspace{0.2em}$ is $\hspace{0.2em} x \hspace{0.2em}$ is replaced with $\hspace{0.2em} 2 \hspace{0.2em}$.

Similarly,

What is the Remainder Theorem?

Consider this. When we divide a number by another, we get a remainder – unless it’s completely divisible. For example, dividing $\hspace{0.2em} 7 \hspace{0.2em}$ by $\hspace{0.2em} 2 \hspace{0.2em}$ gives us $\hspace{0.2em} 3 \hspace{0.2em}$ as the quotient and $\hspace{0.2em} 1 \hspace{0.2em}$ as the remainder.

Similarly, when we divide one polynomial by another, we can have a remainder left over. Here's an example.

So, we end up with $\hspace{0.2em} 6 \hspace{0.2em}$ as the remainder.

Now, would it not be great if we could get the remainder without having to do the whole division?

That’s where the remainder theorem comes in.

It can sound confusing. But it’s not that bad. Let me explain.

Going back to the previous example, here’s the polynomial we were dividing.

And we were dividing it by $\hspace{0.2em} x - 1 \hspace{0.2em}$. So according to the remainder theorem, the remainder would be $\hspace{0.2em} P(1) \hspace{0.2em}$.

The following examples should help make things clearer and remove any doubts you may have.

Examples

Solution

Step 1.  We need to write the divisor as $\hspace{0.2em} x - a \hspace{0.2em}$ so that we can identify what $\hspace{0.2em} a \hspace{0.2em}$ is.

For our present example, the divisor is $\hspace{0.2em} x + 3 \hspace{0.2em}$ and

Comparing $\hspace{0.2em} x - (-3) \hspace{0.2em}$ with $\hspace{0.2em} x - a \hspace{0.2em}$, it's clear that $\hspace{0.2em} a = -3 \hspace{0.2em}$.

Step 2.  Plug the value of $\hspace{0.2em} a \hspace{0.2em}$ obtained from step 1 into the dividend, $\hspace{0.2em} P(x) \hspace{0.2em}$, in place of $\hspace{0.2em} x \hspace{0.2em}$.

And that's it. If we divide $\hspace{0.2em} 4x^2 - 7x - 10 \hspace{0.2em}$ by $\hspace{0.2em} x + 3 \hspace{0.2em}$, the remainder would be $\hspace{0.2em} 47 \hspace{0.2em}$.

Solution

Again, we start by finding the value of $\hspace{0.2em} x \hspace{0.2em}$ for which the divisor becomes $\hspace{0.2em} 0 \hspace{0.2em}$.

Then we plug this value in place of $\hspace{0.2em} x \hspace{0.2em}$ in the dividend.

That’s our remainder.

Solution

Here's the key idea.

So let’s find the remainder for our present case as we did in the previous examples. First, we find the value of $\hspace{0.2em} x \hspace{0.2em}$ for which the divisor becomes $\hspace{0.2em} 0 \hspace{0.2em}$.

Next, we calculate $\hspace{0.2em} P(2) \hspace{0.2em}$ to get the remainder.

So if we divide $\hspace{0.2em} P(x) \hspace{0.2em}$ by $\hspace{0.2em} x - 2 \hspace{0.2em}$, it would leave a remainder of $\hspace{0.2em} -7 \hspace{0.2em}$. That means the given polynomial is not divisible by $\hspace{0.2em} x - 2 \hspace{0.2em}$.

Proof of the Remainder Theorem

If we divide a number, say $\hspace{0.2em} 17 \hspace{0.2em}$, by another number $\hspace{0.2em} (3) \hspace{0.2em}$ we get a quotient $\hspace{0.2em} (5) \hspace{0.2em}$ and a remainder $\hspace{0.2em} (2) \hspace{0.2em}$.

Something similar happens when we divide a polynomial $\hspace{0.2em} P(x) \hspace{0.2em}$ by another polynomial $\hspace{0.2em} (x - c) \hspace{0.2em}$.

Now, let’s see what happens if we substitute $\hspace{0.2em} c \hspace{0.2em}$ in place of $\hspace{0.2em} x \hspace{0.2em}$.

Voila, it gives us the value of the remainder.

We can see that if we divide $\hspace{0.2em} P(x) \hspace{0.2em}$ by $\hspace{0.2em} x - c \hspace{0.2em}$, the remainder would be $\hspace{0.2em} P(c) \hspace{0.2em}$.

And that brings us to the end of this tutorial on the remainder theorem. Until next time.