Dividend is .

Divisor is .

Since the remainder is , is a solution of .

Write the terms of the dividend so that the degrees of the terms are in descending order.

Then write just the coefficients as shown below.

Setup the synthetic division using a zero place for the missing term term in the dividend.

Write the constant of the divisor   to the left.

In this case, , so bring the first coefficient, 1, down.

Multiply the sum, by  : .

Write the product under the next coefficient, and add :.

Multiply the sum, by  : .

Write the product under the next coefficient, and add :.

Multiply the sum, by  : .

Write the product under the next coefficient, and add :.

Multiply the sum, by  : .

Write the product under the next coefficient, and add :.

Multiply the sum, by  : .

Write the product under the next coefficient, and add :.

Multiply the sum,   by  : .

Write the product under the next coefficient, and add :.

The remainder is last entry in the last row.

Therefore, remainder .

The remainder is .

Therefore, when , will have the remainder .

The value of is .