Answer
$x \in \{-3, -3i, 3, 3i\}$ with a multiplicity of $1$
Work Step by Step
The remainder theorem states that when a polynomial $p(x)$ is divided by a linear polynomial $(x - a)$, then the remainder is equal to $p(a)$.
$P(x)=x^4-81$,
factorising the polynomial, $(x^2-9)(x^2+9)=(x-3)(x+3)(x-3i)(x+3i)$
thus, the equation is
$(x-3)(x+3)(x-3i)(x+3i)=0$.
The zeros are:
$x \in \{-3, -3i, 3, 3i\}$ with a multiplicity of $1$.
