Q.Given that $1+2i$ is a root of $x^4-4x^3+6x^2-4x-15=0$, find all the roots.

Theorem. If an equation $f(x)=0$ with real coefficients has a non-real root $a+ib$ ($b\neq0$), then its conjugate $a-ib$ is also a root, with the same multiplicity.

Proof idea. Divide $f(x)$ by the real quadratic $(x-a)^2+b^2=x^2-2ax+(a^2+b^2)$, the minimal real polynomial having $a+ib$ as a root:

for real numbers $R,S$ (the coefficients of $f$ and of the divisor are real, so ordinary long division produces a real quotient and a real linear remainder). Substituting $x=a+ib$ makes the bracketed term vanish by construction, leaving $0=f(a+ib)=R(a+ib)+S=(Ra+S)+iRb$. Since $R,S,a,b$ are real and $b\neq0$, equating real and imaginary parts forces $R=0$ and then $S=0$. So the remainder is identically zero, i.e. $(x-a)^2+b^2$ divides $f(x)$ exactly -- meaning $a-ib$ is a root of $f$ as well, with the same multiplicity as $a+ib$.

Consequence. Non-real roots of a real-coefficient polynomial always occur in conjugate pairs; an odd-degree real equation must therefore always have at least one real root. Once a single non-real root $a+ib$ of an equation is known, its conjugate $a-ib$ is known for free, and dividing out the real quadratic factor $x^2-2ax+(a^2+b^2)$ depresses the equation by two degrees at once.