Problem (Algebra):
For real numbers $\alpha, B$, and $C$, the zeros of $T(x)=x^{3}+x^{2}+B x+C \operatorname{are~}^{2} \alpha$, $\cos ^{2} \alpha$, and $-\csc ^{2} \alpha$. Compute $T(5)$.

Solution:
Use the sum of the roots formula to obtain $\sin ^{2} \alpha+\cos ^{2} \alpha+-\csc ^{2} \alpha=-1$, so $\csc ^{2} \alpha=2$, and $\sin ^{2} \alpha=\frac{1}{2}$. Therefore $\cos ^{2} \alpha=\frac{1}{2}$. T(x) has leading coefficient 1 , so by the factor theorem, $T(x)=\left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)(x+2)$. Then $T(5)=\left(5-\frac{1}{2}\right)\left(5-\frac{1}{2}\right)(5+2)=\frac{567}{4}$.

Final answer: $\frac{567}{4}$