The remainder theorem states – if you divide a polynomial $\hspace{0.2em} P(x) \hspace{0.2em}$ by $\hspace{0.2em} x - a \hspace{0.2em}$, the remainder would be $\hspace{0.2em} P(a) \hspace{0.2em}$.

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.

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

$P(x) \, = \, 2x^2 + x - 1$

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 $\hspace{0.2em} x \hspace{0.2em}$.

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 $\hspace{0.2em} 2 \hspace{0.2em}$.

$\begin{align*} P( {\color{Red} 2} ) \, &= \, 2 \cdot {\color{Red} 2} ^2 + {\color{Red} 2} - 1 \\[1em] &= \, 9 \end{align*}$

Similarly,

$\begin{align*} P( {\color{Red} 0} ) \, &= \, 2 \cdot {\color{Red} 0} ^2 + {\color{Red} 0} - 1 \\[1em] &= \, -1 \end{align*}$

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.

$P(x) \, = \, x^2 + x + 4$

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

$\begin{align*} P( {\color{Red} 1} ) \, &= \, {\color{Red} 1} ^2 + {\color{Red} 1} + 4 \\[1em] &= \, 6 \end{align*}$

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

Example 1

What would be the remainder if we divide $\hspace{0.2em} P(x): 4x^2 - 7x - 10 \hspace{0.2em}$ by $\hspace{0.2em} x + 3\hspace{0.10em} \hspace{0.2em}$?

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

$x + 3 = x - (-3)$

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

Quick Tip – Probably the simplest to find $\hspace{0.2em} a \hspace{0.2em}$ is as follows. We equate the divisor to $\hspace{0.2em} 0 \hspace{0.2em}$ and solve for $\hspace{0.2em} x \hspace{0.2em}$. The solution would be $\hspace{0.2em} a \hspace{0.2em}$

$\begin{align*} x + 3 \, &= \, 0 \\[1em] x \, &= \, -3 \end{align*}$

So again, $\hspace{0.2em} a = -3 \hspace{0.2em}$.

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 $\hspace{0.2em} x \hspace{0.2em}$.

$\begin{align*} P( {\color{Red} -3} ) \, &= \, 4( {\color{Red} -3} )^2 - 7 \cdot {\color{Red} -3} - 10 \\[1em] &= \, 47 \end{align*}$

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 $\hspace{0.2em} 47 \hspace{0.2em}$.

Example 1

What would be the remainder if we divide $\hspace{0.2em} x^2 + 2x - 3 \hspace{0.2em}$ by $\hspace{0.2em} x - 1 \hspace{0.2em}$?

Solution

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

$\begin{align*} x - 1 \, &= \, 0 \\[1em] x \, &= \, 1 \end{align*}$

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

${\color{Red} 1} ^2 + 2 \cdot {\color{Red} 1} - 3 \, = \, 0$

That’s our remainder.

A remainder of zero implies that $\hspace{0.2em} x - 1 \hspace{0.2em}$ is a factor of $\hspace{0.2em} x^2 + 2x - 3 \hspace{0.2em}$.

Example 3

Is the polynomial $\hspace{0.2em} P(x): x^2 - 3x - 5 \hspace{0.2em}$ evenly divisible by $\hspace{0.2em} x - 2 \hspace{0.2em}$?

Solution

Here's the key idea.

If a polynomial is evenly divisible by another, the remainder would be $\hspace{0.2em} 0 \hspace{0.2em}$. And vice-versa.

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 $\hspace{0.2em} 0 \hspace{0.2em}$.

$\begin{align*} x - 2 \, &= \, 0 \\[1em] x \, &= \, 2 \end{align*}$

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

$\begin{align*} P( {\color{Red} 2} ) \, &= \, {\color{Red} 2} ^2 - 3 \cdot {\color{Red} 2} - 5 \\[1em] &= \, -7 \end{align*}$

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 $\hspace{0.2em} x - 2 \hspace{0.2em}$.

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 $\hspace{0.2em} (x - c) \hspace{0.2em}$.

In the case of polynomials, the degree of the remainder would always be less than that of the divisor. So if the divisor is of the first degree (of the form $\hspace{0.2em} x - c \hspace{0.2em}$), the remainder would have a degree $\hspace{0.2em} 0$. That means it would constant (usually denoted by $\hspace{0.2em} r \hspace{0.2em}$).

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

$\begin{align*} P( {\color{Red} c} ) \, &= \, (c - {\color{Red} c} ) \cdot Q( {\color{Red} c} )+ r \\[1em] P( {\color{Red} c} ) \, &= \, 0 \cdot Q( {\color{Red} c} )+ r \\[1em] P( {\color{Red} c} ) \, &= \, r \end{align*}$

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 $\hspace{0.2em} P(c) \hspace{0.2em}$.

If $\hspace{0.2em} x - c \hspace{0.2em}$ is a factor of a polynomial $\hspace{0.2em} P(x) \hspace{0.2em}$, the remainder upon division would be $\hspace{0.2em} 0 \hspace{0.2em}$. And vice-versa. But we also know – from the remainder theorem - that the remainder would be $\hspace{0.2em} P(c) \hspace{0.2em}$.

So, in other words, $\hspace{0.2em} x - c \hspace{0.2em}$ is a factor of $\hspace{0.2em} P(x) \hspace{0.2em}$ if and only if $\hspace{0.2em} P(c) = 0 \hspace{0.2em}$. And that's the factor theorem.

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

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