Question

Use the Remainder Theorem to find the reminder when P(x)=x3+4x2+1 is divided by (i+1)x+2.

Answer 1

First, rewriting a bit,

`(x^3+4x^2+1)/((i+1)x+2) = \frac1{i+1} \cdot \frac{x^3+4x^2+1}{x+\frac2{i+1}}`

By the remainder theorem, the remainder upon dividing
`x^3+4x^2+1` by
`x+\frac2{i+1}` is equal to

`\left(-\frac2{i+1}\right)^3 + 4\left(-\frac2{i+1}\right)^2 + 1 = \boxed{3 - 6i}`

which is to say,

`\frac1{i+1} \cdot \frac{x^3+4x^2+1}{x+\frac2{i+1}} = q(x) + \frac{3-6i}{(i+1)\left(x+\frac2{i+1}\right)}= q(x) + (3-6i)/((i+1)x+2)`