The remainder and factor theorems are two algebraic theorems related to the division and factorization of polynomials. The remainder theorem allows us to find the remainder when we divide a polynomial by a linear expression. On the other hand, the factor theorem allows us to verify the factors of a polynomial.

Here, we will learn everything related to the remainder and factor theorems. We will look at their statements, the proofs of the theorems, and some examples.

## What is the remainder theorem?

The remainder theorem is an algebraic theorem stating that when we divide a polynomial $latex f(x)$ by $latex (\alpha x-\beta)$, the remainder is $latex f\left(\frac{\beta }{\alpha}\right)$.

This means that we can use the Remainder Theorem to find the remainder when dividing a polynomial by the expression $latex (\alpha x-\beta)$.

### Proof of the remainder theorem

The result of a division is the quotient. Therefore, if we multiply the quotient by the divisor and add the remainder, we get the original polynomial. That is, we have:

$$f(x)=(\alpha x-\beta)(\text{Quotient})+(\text{Remainder})$$

Now, we are going to use the value $latex x=\frac{\beta}{\alpha}$, to obtain the following:

$$f\left(\frac{\beta}{\alpha}\right)=\left[ \alpha \left(\frac{\beta}{\alpha}\right)-\beta\right](\text{Quotient})+(\text{Remainder})$$

$$=\left[ \beta-\beta\right](\text{Quotient})+(\text{Remainder})$$

$$=\left[ 0 \right](\text{Quotient})+(\text{Remainder})$$

$$=\text{Remainder}$$

We see that the Remainder Theorem is true since by using $latex x=\frac{\beta}{\alpha}$ in the function, we get the remainder of the division.

## What is the factor theorem?

The factor theorem is an algebraic theorem that tells us that if $latex (\alpha x-\beta)$ is a factor of the polynomial $latex f(x)$, then $latex f\left(\frac{ \beta}{\alpha}\right)=0$.

This means that the factor theorem connects the factors and the zeros of a polynomial. The remainder theorem can be considered as a specific version of this theorem.

Additionally, we can use the factor theorem to remove known zeros from a polynomial and leave all unknown zeros intact, resulting in a lower degree polynomial with easier zeros to find.

### Proof of the factor theorem

If $latex (\alpha x-\beta)$ is a factor of $latex f(x)$, then the remainder equals 0 when we divide $latex f(x)$ by $latex (\alpha x -\beta)$.

Therefore, if we use the Remainder Theorem, we can show that

$$f\left(\frac{\beta}{\alpha}\right)=0$$

## Examples of the remainder and factor theorems

The following examples are solved by applying the remainder and factor theorems. Each example has a detailed solution.

**EXAMPLE 1**

Find the remainder when we divide the polynomial $latex x^3+5x^2-17x-21$ by $latex x-4$.

**Solution:** To solve this, we have to use the Remainder Theorem.

The Remainder Theorem tells us that when $latex f(x)$ is divided by $latex x-4$, it is $latex f(4)$. Therefore, we have:

$$f(4)=(4)^3+5(4)^2-17(4)-21$$

$latex =64+80-68-21$

$latex =55$

The remainder of the division is 55.

**EXAMPLE **2

**EXAMPLE**2

When we divide the polynomial $latex 3x^3+bx^2-7x+5$ by $latex x+3$, the remainder is 17. Find the value of *b*.

**Solution:** In this case, we are going to use the Remainder Theorem to form an equation and then solve for *b*.

Using the remainder theorem, we know that $latex f(-3)=17$. Thus, we have:

$$3(-3)^3+b(-3)^2-7(-3)+5=17$$

$latex -81+9b+21+5=17$

$latex 9b=72$

$latex b=8$

The value of *b* is 8.

**EXAMPLE **3

**EXAMPLE**

Determine if $latex (x+1)$ is a factor of the polynomial $latex f(x) = {x}^2 + x – 2$.

**Solution:** To determine if $latex (x+1)$ is a factor, we are going to use the factor theorem.

The factor theorem tells us that if *x*+1 is a factor of the polynomial, we must have $latex f(-1)=0$. Therefore, using $latex x=-1$ in the polynomial, we have:

$latex f(x) = {x}^2 – x -2$

$latex f(-1) = {(-1)}^2 – (-1) -2$

$latex f(-1) = 1 +1 – 2 $

$latex f(-1) = 0$

We see that *f*(-1) = 0. This means that $latex (x + 2)$ is a factor of the given polynomial.

**EXAMPLE **4

**EXAMPLE**

Is $latex (2x-1)$ a factor of the polynomial $latex f(x) = 2{x}^2 – x – 1$?

**Solution:** Again, we are going to use the factor theorem to solve this.

If 2*x*-1 is a factor of the polynomial, we should have $latex f(\frac{1}{2})=0$. Substituting the value $latex \frac{1}{2}$ into the polynomial, we have:

$latex f(\frac{1}{2}) = 2{x}^2 – x – 1$

$latex f(\frac{1}{2}) = 2{(\frac{1}{2})}^2 – (\frac{1}{2}) – 1$

$latex f(\frac{1}{2}) = 2(\frac{1}{4}) – (\frac{1}{2}) – 1$

$latex f(\frac{1}{2}) = \frac{1}{2} – (\frac{1}{2}) – 1$

We got $latex f(\frac{1}{2}) = -1$. This means that (2x–1) is not a factor of $latex f(x) = 2{x}^2 – x – 1$.

## See also

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