Q.Find all zeros of the polynomial $x^6-3x^5-5x^4+22x^3-39x^2-39x+135$, if it is known that $1+2i$ and $\sqrt3$ are two of its zeros.

Step 1. Identify the automatic extra roots. Since coefficients are real, $1+2i$ a root forces $1-2i$ a root too; since coefficients are rational, $\sqrt3$ a root forces $-\sqrt3$ a root too.

Step 2. Build the two quadratic factors. $(x-(1+2i))(x-(1-2i))=(x-1)^2+4=x^2-2x+5$. $(x-\sqrt3)(x+\sqrt3)=x^2-3$.

Step 3. Multiply them into one quartic factor. $(x^2-2x+5)(x^2-3)=x^4-2x^3+2x^2+6x-15$.

Step 4. Divide the given sextic by this quartic. $x^6-3x^5-5x^4+22x^3-39x^2-39x+135 \div (x^4-2x^3+2x^2+6x-15)$ gives quotient $x^2-x-9$ with remainder $0$ (verified by direct expansion, matching term by term).

Step 5. Solve the remaining quadratic. $x^2-x-9=0 \implies x=\dfrac{1\pm\sqrt{1+36}}2=\dfrac{1\pm\sqrt{37}}2$. …

Complex Conjugate Root Theorem combined with Irrational Conjugate Roots — build and divide out the degree-4 factor from the two known roots, then solv …

• Forgetting $\sqrt3$ ALSO needs a rational-coefficients conjugate partner $-\sqrt3$, just like the imaginary root needs its conjugate …