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# Remainder Theorem

Remainders are a familiar concept when dividing numbers, but did you know that a similar concept, known as the Remainder Theorem, can be applied to polynomials? The Remainder Theorem is a powerful tool that helps us understand and solve a wide range of polynomial functions without resorting to long division or synthetic division. In this section, we will explore the principles behind the Remainder Theorem and how it can provide valuable insights into the behavior of polynomial functions. So, let's delve into the world of the Remainder Theorem and discover how it can enhance our understanding of mathematics.

## The Remainder Theorem

The Remainder Theorem states that when a polynomial $f\left(x\right)$ is divided by a binomial of the form $\left(x-c\right)$ , where c is a constant, the remainder can be found by evaluating the polynomial at $x=c$

x = c

. In other words, the remainder is equal to $f\left(c\right)$ .

To clarify with an example, let's say we have the following polynomial:

$f\left(x\right)=5{x}^{3}-3{x}^{2}+5x-3$

And we want to divide it by the binomial $\left(x-2\right)$ . According to the remainder theorem, we can find the remainder by evaluating the polynomial at $x=2$ :

$f\left(2\right)=5{\left(2\right)}^{3}-3{\left(2\right)}^{2}+5\left(2\right)-3$

$f\left(2\right)=5\left(8\right)-3\left(4\right)+10-3$

$f\left(2\right)=40-12+10-3$

$f\left(2\right)=35$

So, when we divide the polynomial $f\left(x\right)$ by the binomial $\left(x-2\right)$ , the remainder is 35.

## The Remainder Theorem

We can divide polynomials using synthetic division, a quicker method than long division when dividing by a binomial of the form $\left(x-c\right)$ , where c is a constant.

For example, let's divide the polynomial $f\left(x\right)={x}^{2}-8x+6$ by the binomial $x-2$ :

Write only the coefficients of the dividend inside the box and the constant c from the divisor $\left(x-c\right)$ outside the top left of the box.

Bring the first coefficient inside the box down unchanged to the bottom of the box.

Multiply the carried-down value by the constant c from the divisor, and write the result below the second coefficient of the dividend.

Add the value obtained in step 3 to the second coefficient of the dividend and write the sum at the bottom of the box.

Repeat steps 3 and 4 for the remaining coefficients of the dividend.

The last value obtained in the process is the remainder. The other values at the bottom of the box represent the coefficients of the quotient.

In our example, after performing synthetic division, we get the following result:

The remainder is -6, and the quotient is $x-6$ .

Remember to add back the variables to the coefficients of the quotient to express the result correctly.

## You don't need to divide to find the remainder

We don't really need to divide in order to find the remainder.

If we have a function that looks like this:

$\frac{f\left(x\right)}{\left(x-c\right)}$

We know that the remainder is always $f\left(c\right)$ .

For example, $\frac{2{x}^{2}-5x-1}{x-3}$ tells us that the remainder is $f\left(3\right)$ .

$f\left(3\right)=2{\left(3\right)}^{2}-5\left(3\right)-1$

$=2×9-5×3-1$

$=18-15-1$

$=2$

## Flashcards covering the Remainder Theorem

Algebra II Flashcards

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