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,

$\dfrac{x^3+4x^2+1}{(i+1)x+2} = \dfrac1{i+1} \cdot \dfrac{x^3+4x^2+1}{x+\frac2{i+1}}$

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

$\left(-\dfrac2{i+1}\right)^3 + 4\left(-\dfrac2{i+1}\right)^2 + 1 = \boxed{3 - 6i}$

which is to say,

$\dfrac1{i+1} \cdot \dfrac{x^3+4x^2+1}{x+\frac2{i+1}} = q(x) + \dfrac{3-6i}{(i+1)\left(x+\frac2{i+1}\right)}= q(x) + \dfrac{3-6i}{(i+1)x+2}$