Polynomials: Remainder theorem

Write $x^{2025}=q(x)(x^2+1)(x^2+x+1)+r(x)$ , where $r(x)$ is the remainder.

Substituting $x=i$ gives $i^{2025}=i=r(i)$ . From the options, only the third option satisfies $r(i)=i$ .

For a rigorous solution, let $r(x)=ax^3+bx^2+cx+d$ , and plug all the roots of $(x^2+1)(x^2+x+1)$ one by one to get the equations and solve for the coefficients.

By remainder theorem , we have:

For a polynomial $f(x)$ , the remainder of $f(x)$ upon division by $x-c$ is $f(c)$ .

Let the remainder of $x^{2025}$ upon division by $(x^2+1)(x^2 + x + 1)$ be $r(x) = x^3 + ax^2 + bx + c$ . We note that the roots of $x^2 + 1$ are $\pm i$ and those of $x^2 + x + 1$ is the cubic root of unity $\omega$ . Therefore we have:

$\begin{aligned} r(i) & = f(i) \\ i^3 + ai^2 + bi + c & = i^{2025} \\ -i - a + bi + c & = i \end{aligned}$

Equating the real and imaginary parts on both sides: $\begin{cases} -a+c = 0 & \implies a = c \\ -1 + b = 1 & \implies b = 2 \end{cases}$ . We get the same results using $-i$ instead of $i$ .

And we have:

$\begin{aligned} r(\omega) & = f(\omega) \\ \omega^3 + a\omega^2 + b\omega + c & = \omega^{2025} & \small \blue{\text{Note that }\omega^3 = 1} \\ 1 + a\omega^2 + 2\omega + a & = 1 \\ a\omega^2 + 2\omega + a & = 0 & \small \blue{\text{and that }\omega^2 + \omega + 1 = 0} \\ \implies a = c & = 2 \end{aligned}$

Therefore $r(x) = \boxed{x^3 + 2x^2 + 2x + 2}$ .