Factor Theorem

In this tutorial, we'll look at the factor theorem — what it states and how one can use it. But before we dive into that, let’s get the basics out of the way.

Polynomials and Their Notation

Here’s a trinomial, a polynomial with three terms.

9x^2 - x + 4

Now, polynomials are generally denoted using a letter from the English alphabet, often $\hspace{0.2em} P \hspace{0.2em}$ or $\hspace{0.2em} Q \hspace{0.2em}$ . For example,

P(x) \, = \, 9x^2 - x + 4

The $\hspace{0.2em} x \hspace{0.2em}$ inside parentheses indicates that the variable in the polynomial is $\hspace{0.2em} x \hspace{0.2em}$ .

Value of a Polynomial

Say, we have a polynomial $\hspace{0.2em} P(x) \hspace{0.2em}$ . Now, if we replace $\hspace{0.2em} x \hspace{0.2em}$ with some constant $\hspace{0.2em} {\color{Red} a} \hspace{0.2em}$ , the resulting value is denoted as $\hspace{0.2em} P( {\color{Red} a} ) \hspace{0.2em}$ .

So, $\hspace{0.2em} P( {\color{Red} 1} ) \hspace{0.2em}$ refers to the value of the polynomial $\hspace{0.2em} P \hspace{0.2em}$ if $\hspace{0.2em} x \hspace{0.2em}$ is replaced with $\hspace{0.2em} {\color{Red} 1} \hspace{0.2em}$ . For the polynomial above, that would mean —

\begin{align*} P( {\color{Red} 1} ) \, &= \, 9 \cdot {\color{Red} 1} ^2 - {\color{Red} 1} + 4 \\[1em] &= \, 12 \end{align*}

Similarly,

\begin{align*} P( {\color{Red} 2} ) \, &= \, 9 \cdot {\color{Red} 2} ^2 - {\color{Red} 2} + 4 \\[1em] &= \, 38 \end{align*}

What Is the Factor Theorem?

The factor theorem states that for any polynomial $\hspace{0.2em} P(x) \hspace{0.2em}$ ,

if $\hspace{0.2em} P(a) = 0 \hspace{0.2em}$ , $\hspace{0.2em} x - a \hspace{0.2em}$ is a factor of $\hspace{0.2em} P(x) \hspace{0.2em}$ , and
if $\hspace{0.2em} x - a \hspace{0.2em}$ is a factor of $\hspace{0.2em} P(x) \hspace{0.2em}$ , $\hspace{0.2em} P(a) = 0 \hspace{0.2em}$

Confused? No worries. Let me explain.

Here’s a polynomial,

P(x) \, = \, x^2 - 3x + 2

Understanding the first part

Let’s find the value of the polynomial for $\hspace{0.2em} x = 1 \hspace{0.2em}$ .

\begin{align*} P( {\color{Red} 1} ) \, &= \, {\color{Red} 1} ^2 - 3 \cdot {\color{Red} 1} + 2 \\[1em] &= \, 0 \end{align*}

Now, according to the factor theorem (its first part), because our polynomial reduces to $\hspace{0.2em} 0 \hspace{0.2em}$ for $\hspace{0.2em} x = 1 \hspace{0.2em}$ , $\hspace{0.2em} x - 1 \hspace{0.2em}$ must be a factor of the polynomial.

So, let’s verify that.

No remainders. So $\hspace{0.2em} x - 1 \hspace{0.2em}$ is indeed a factor of the polynomial.

Understanding the second part

Great. But what if we knew that $\hspace{0.2em} x - 2 \hspace{0.2em}$ is a factor of $\hspace{0.2em} P(x) \hspace{0.2em}$ ? Can the factor theorem give us some more information about the polynomial? Yes, that’s where the second part of the theorem comes in.

According to the factor theorem, if $\hspace{0.2em} x - 2 \hspace{0.2em}$ is a factor of the polynomial, then $\hspace{0.2em} P(2) \hspace{0.2em}$ must be equal to $\hspace{0.2em} 0 \hspace{0.2em}$ . Let’s check.

\begin{align*} P( {\color{Red} 2} ) \, &= \, {\color{Red} 2} ^2 - 3 \cdot {\color{Red} 2} + 2 \\[1em] &= \, 0 \end{align*}

And the polynomial reduces to $\hspace{0.2em} 0 \hspace{0.2em}$ . That means $\hspace{0.2em} x - 2 \hspace{0.2em}$ is a factor too.

Using the Factor Theorem – Examples

Example 1

Are $\hspace{0.2em} \hspace{0.25em} x + 1\hspace{0.25em} \hspace{0.2em}$ and/or $\hspace{0.2em} \hspace{0.25em} x - 3\hspace{0.25em} \hspace{0.2em}$ factors of $\hspace{0.2em} \hspace{0.25em} P(x): x^2 + 2x - 15\hspace{0.25em} \hspace{0.2em}$ ?

Solution

Let's check for $\hspace{0.2em} x + 1 \hspace{0.2em}$ first.

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 $\hspace{0.2em} x + 1 \hspace{0.2em}$ and

x + 1 = x - (-1)

Comparing $\hspace{0.2em} x - (-1) \hspace{0.2em}$ with $\hspace{0.2em} x - a \hspace{0.2em}$ , it's clear that $\hspace{0.2em} a = -1 \hspace{0.2em}$ .

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

So again, $\hspace{0.2em} \hspace{0.25em}a = -1 \hspace{0.2em}$ .

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(-1) \, &= \, ( {\color{Red} -1} )^2 + 2 \cdot {\color{Red} -1} - 15 \\[1em] &= \, -16 \end{align*}

Step 3. Since the resulting value of the polynomial in step 2 is non-zero, $\hspace{0.2em} \hspace{0.25em}x + 1\hspace{0.25em} \hspace{0.2em}$ is not a factor of $\hspace{0.2em} \hspace{0.25em}x^2 + 2x – 15\hspace{0.15em} \hspace{0.2em}$ .

Alright, now we can move to $\hspace{0.2em} x - 3 \hspace{0.2em}$ .

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

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

Next, we plug the value obtained into $\hspace{0.2em} P(x) \hspace{0.2em}$ . Meaning, we calculate $\hspace{0.2em} P(3) \hspace{0.2em}$ .

\begin{align*} P(3) \, &= \, {\color{Red} 3} ^2 + 2 \cdot {\color{Red} 3} - 15 \\[1em] &= \, 0 \end{align*}

Well, this time the polynomial reduces to $\hspace{0.2em} 0 \hspace{0.2em}$ . So, $\hspace{0.2em} \hspace{0.25em}x - 3\hspace{0.25em} \hspace{0.2em}$ is a factor of $\hspace{0.2em} \hspace{0.25em}P(x)\hspace{0.15em} \hspace{0.2em}$ .

Example 2

Using the factor theorem only, find one factor of $\hspace{0.2em} x^2 + x - 6 \hspace{0.2em}$

Solution

Before anything else, let's give the polynomial a name. It would make it easier to talk about it. So,

P(x) = x^2 + x - 6

Now, in order to find factors of a polynomial using the factor theorem, we have the trial and error method.

In this method, we calculate the value of the polynomial for different values of $\hspace{0.2em} x \hspace{0.2em}$ , until the polynomial reduces to $\hspace{0.2em} 0 \hspace{0.2em}$ . We can repeat the process to find more factors if needed (and if they exist).

Quick Tip — As long as all the coefficients in the polynomial are integers, you can limit your search to the factors (positive and negative) of the constant term.

For example, here the constant term is $\hspace{0.2em} -6 \hspace{0.2em}$ . So, we need to try its factors $\hspace{0.2em} (1 \hspace{0.2em}$ , $\hspace{0.2em} -1 \hspace{0.2em}$ , $\hspace{0.2em} 2 \hspace{0.2em}$ , $\hspace{0.2em} -2 \hspace{0.2em}$ , $\hspace{0.2em} 3 \hspace{0.2em}$ , and $\hspace{0.2em} -3) \hspace{0.2em}$ only.

Alright, let’s try $\hspace{0.2em} 1 \hspace{0.2em}$ first.

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

The result is non-zero. So, $\hspace{0.2em} \hspace{0.25em}x - 1\hspace{0.25em} \hspace{0.2em}$ is not a factor.

Okay, what about $\hspace{0.2em} -1 \hspace{0.2em}$ ?

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

Again, doesn’t work. So, $\hspace{0.2em} \hspace{0.25em}x - (-1)\hspace{0.25em} \hspace{0.2em}$ or $\hspace{0.2em} \hspace{0.25em}x + 1\hspace{0.25em} \hspace{0.2em}$ is also not a factor. We move to the next one.

\begin{align*} P( {\color{Red} 2} ) \, &= \, {\color{Red} 2} ^2 + {\color{Red} 2} - 6 \\[1em] &= \, 0 \end{align*}

Yes. The polynomial reduces to $\hspace{0.2em} 0 \hspace{0.2em}$ . so $\hspace{0.2em} \hspace{0.25em}x - 2\hspace{0.25em} \hspace{0.2em}$ is a factor of $\hspace{0.2em} \hspace{0.25em}x^2 + x - 6 \hspace{0.2em}$ . Job done.

Proof of the Factor Theorem

If we divide a number, say $\hspace{0.2em} 19 \hspace{0.2em}$ , by another number $\hspace{0.2em} (7) \hspace{0.2em}$ we get a quotient $\hspace{0.2em} (2) \hspace{0.2em}$ and a remainder $\hspace{0.2em} (5) \hspace{0.2em}$ .

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

Factor theorem - divisor, quotient, and remainder demo

Using this last expression, we’ll prove the factor theorem in two parts.

Part $\hspace{0.2em} - \hspace{0.2em} 1 \hspace{0.2em}| \hspace{0.2em}$ If $\hspace{0.2em} x - c \hspace{0.2em}$ is a factor of $\hspace{0.2em} P(x) \hspace{0.2em}$ , $\hspace{0.2em} P(c) = 0 \hspace{0.2em}$ .

We start by assuming that $\hspace{0.2em} x - c \hspace{0.2em}$ is a factor of $\hspace{0.2em} P(x) \hspace{0.2em}$ . That means there would be no remainder if we do the division. So –

P(x) \, = \, (x - c) \cdot Q(x)

Substituting $\hspace{0.2em} c \hspace{0.2em}$ in place of $\hspace{0.2em} x \hspace{0.2em}$ , we get

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

Hence proved.

Part $\hspace{0.2em} - \hspace{0.2em} 2 \hspace{0.2em}| \hspace{0.2em}$ If $\hspace{0.2em} P(c) = 0 \hspace{0.2em}$ , $\hspace{0.2em} x - c \hspace{0.2em}$ is a factor of $\hspace{0.2em} P(x) \hspace{0.2em}$ .

This time we assume $\hspace{0.2em} P(c) = 0 \hspace{0.2em}$ , and start with this expression from above.

P(x) \, = \, (x - c) \cdot Q(x) + r

Plugging in $\hspace{0.2em} c \hspace{0.2em}$ in place of $\hspace{0.2em} x \hspace{0.2em}$ , we have

P( {\color{Red} c} ) \, = \, ( {\color{Red} c} - c) \cdot Q( {\color{Red} c} ) + r

Now, since $\hspace{0.2em} P(c) = 0 \hspace{0.2em}$ and $\hspace{0.2em} c - c = 0 \hspace{0.2em}$ , of course –

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

Again, proved.

Combining parts $\hspace{0.2em} 1 \hspace{0.2em}$ and $\hspace{0.2em} 2 \hspace{0.2em}$ , we have proved the factor theorem.

And with that, we come to the end of this tutorial on the factor theorem. This might be a great time for you to take a quick look at a closely related theorem, the remainder theorem.

Until next time.