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)#?

#f(x) = x^4 - x^3 -5x^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(1) != 0# so # (x-1)# is not a factor
#f(-1) != 0# so # (x+1)# is not a factor
#f(-2) = 0# so #(x+2)# is a factor. etc..

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

#f(+3) =3^4 - 3^3 -5(3)^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 " ) 1 -1 -5 -1 -6"#
#" 3 6 3 6"#

#" 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#