If $\alpha$ and $\beta$ are the roots of $x^2-4x+5=0$, find $(\alpha^3-4\alpha^2+6\alpha+2)(\beta^3-4\beta^2+6\beta+2)$

Question

If $\alpha$ and $\beta$ are two roots of the equation $x^2-4x+5=0$, find $$(\alpha^3 - 4\alpha^2 + 6\alpha + 2)(\beta^3 - 4\beta^2 + 6\beta + 2)$$ using Vieta's relations.

Answer 1

$$ \left( x^{3} - 4 x^{2} + 6 x + 2 \right) = \left( x^{2} - 4 x + 5 \right) \cdot \left( x \right) + \left( x + 2 \right) $$

This means that your product is just $$ (\alpha+2)(\beta+2) = (\alpha \beta) + 2(\alpha + \beta) + 4 $$

which you should now be able to evaluate