8. Let $x_{1}, x_{2}$ be the roots of the equation $x^{2}-x-4=0$. Find $\left(x_{1}^{5}-20 x_{1}\right) \cdot\left(x_{2}^{4}+16\right)$.
Answer 76

Solution: If $x$ is one of the roots of the equation, then $x^{2}=x+4$. Squaring, we get $x^{4}=$ $x^{2}+8 x+16=9 x+20$. Multiplying by $x$, we get $x^{5}=9 x^{2}+20 x=29 x+36$. Substituting $x$ with the roots of the equation, we get: $\left(x_{1}^{5}-20 x_{1}\right) \cdot\left(x_{2}^{4}+16\right) .=\left(29 x_{1}+36-20 x_{1}\right) \cdot\left(9 x_{2}+16\right)=$ $\left(9 x_{1}+16\right)\left(9 x_{2}+16\right)=81 x_{1} x_{2}+144\left(x_{1}+x_{2}\right)+256$. By Vieta's theorem $x_{1} x_{2}=-4, x_{1}+$ $x_{2}=1$. Substituting, we get $81 *(-4)+144+256=76$

All problems were scored out of 15 points.