If $\alpha $ is a root for $ x^2-2x+3=0 $, Proove that $\alpha^2-2\alpha^3=9 $
I have tried the following,
since $\alpha$ is a root, it should satisfy the equation. Hence, $$\alpha^2-2\alpha+3=0 $$
Since this equation has complex roots, other root should be the conjugate of $\alpha$, Do I need to consider this? This looks so simple but cannot work it out further. Please help. Thanks!
We have $\alpha^2=2\alpha-3$.
$$\alpha^2-2\alpha^3=\alpha^2(1-2\alpha)=(2\alpha-3)(1-2\alpha)=-4\alpha^2+8\alpha-3=-4(2\alpha-3)+8\alpha-3=9$$
Hint: $\;\alpha^2=2 \alpha - 3\,$, so $\,\alpha^2 - 2 \alpha^2 \alpha = (2\alpha-3)-2(2\alpha-3)\alpha=-4\alpha^2+8\alpha-3\ldots\,$
$$ \left( 2 x^{3} - x^{2} + 9 \right) = \left( x^{2} - 2 x + 3 \right) \cdot \left( 2 x + 3 \right) $$
