When can a conjugate root be assumed?

The source explanation is that $p$ may not be real: the question has not yet stated or established that every coefficient is real. Therefore the conjugate-root theorem cannot yet be invoked on that premise. This concerns the justification at this stage; part (ii) will establish that this particular p is real.

Substitute $1-2i$. Its square is $-3-4i$, its cube is $-11+2i$ and its fourth power is $-7+24i$. Thus $(-7+24i)-(-11+2i)+4(-3-4i)+3-6i+p=0$, giving $-5+p=0$. Hence $p=5$.

Now all coefficients are real, so $1+2i$ is a root. The conjugate factors multiply to $(z-1-2i)(z-1+2i)=(z-1)^2-(2i)^2=z^2-2z+5$. Write the remaining factor as $z^2+Az+B$. The constant coefficient gives $B=1$; the cubic coefficient gives $A-2=-1$, so $A=1$. Hence $(z^2-2z+5)(z^2+z+1)=0$. Solving $z^2+z+1=0$ gives $(-1\pm i\sqrt3)/2$. The other roots are $1+2i$, $(-1+i\sqrt3)/2$ and $(-1-i\sqrt3)/2$.