# How do you use the remainder theorem and synthetic division to find the remainder when (x^4-x^3-5x^2-x-6) div (x-3)?

Jun 22, 2016

We use the remainder theorem to determine whether (x-3) is a factor of the expression. The remainder is 0. Therefore it is a factor and we can use synthetic division to find the quotient.
(x^4 - x^3 -5x^2 -x -6) ÷ (x-3) = x^3 + 2x^2 + x + 2

#### Explanation:

$f \left(x\right) = {x}^{4} - {x}^{3} - 5 {x}^{2} - x - 6$

The factors of 6 are: 1, -1, 2, -2, 3, -3, 6, -6.
Substitute each into f(x) until you obtain a result of 0.

$f \left(1\right) \ne 0$ so $\left(x - 1\right)$ is not a factor
$f \left(- 1\right) \ne 0$ so $\left(x + 1\right)$ is not a factor
$f \left(- 2\right) = 0$ so $\left(x + 2\right)$ is a factor. etc..

However, we are asked to find the remainder when the expression is divided by $\left(x - 3\right) .$ Substitute $x = 3$ into the expression.

$f \left(+ 3\right) = {3}^{4} - {3}^{3} - 5 {\left(3\right)}^{2} - 3 - 6 = 0$.
This means there is no remainder and (x-3) is a factor.

By synthetic division:

Write down only the coefficients of the terms, and write +3 outside.

$3 \text{ ) 1 -1 -5 -1 -6}$
$\text{ 3 6 3 6}$

$\text{ 1 2 1 2 0}$

Bring down the first 1.
Multiply it by 3 and write the answer under the second number. Add.
This gives 2.

Multiply 3 by 2 and write the answer under the third number. Add.
This gives 1.
Multiply 3 by the 1 and write it under the fourth number. Add.
This gives 2.
Multiply 3 by 2 and write it under the fifth number. Add.

This gives 0. Which means there is no remainder and (x-3) is a factor of the expression.

The quotient will start with an ${x}^{3}$ term because x^4 ÷ x = x^3

Use the numbers in last row as the coefficients of the terms with the descending powers of x.

(x^4 - x^3 -5x^2 -x -6) ÷ (x-3) = x^3 + 2x^2 + x + 2