Question

If R(x) is the remainder when(x^2019)– 1 is divided by (x² + 1)(x² + x + 1). Then what is the value of R(4)?

Answer 1

Given: R(x) is the remainder when(x^2019)– 1 is divided by (x² + 1)(x² + x + 1).

To find: What is the value of R(4)?

Solution:

• Now we have given (x^2019) - 1 is divided by (x² + 1)(x² + x + 1).

• Let the divisor be d, and remainder be ax + b.

• So the equation will be:

                  ( x^2019 ) - 1 = d × (x² + 1)(x² + x + 1) + ax + b

• Let x = 0, we get:

                  -1 = d + b

                  b = -1-d   .......................(i)

• Let x = 1, we get:

                  0 = 6d + a + b ..............(ii)

• Let x = -1, we get:

                  -2 = 2d - a + b  ................(iii)

• Putting i in iii, we get:

                  -2 = 2d - a -1 - d

                  -1 = d - a

                  a = d + 1 .....................(iv)

• Putting i and iv in ii, we get:

                  0 = 6d + d + 1 - 1 - d

                  0 = 6d

                  d = 0

• Putting d = 0 in i and iv, we get:

                  b = -1

                  a = 1

• So the remainder will be: R(x) = ax + b = x - 1

                  R(4) = 4-1 = 3

Answer:

               So the value of R(4) is 3.