6. Three of the roots of $x^{4}+a x^{2}+b x+c=0$ are 3, -2, and 5. Find the value of $a+b+c$.
Answer: 167
Solution: The coefficient of $x^{3}$ is 0 , so by Vieta's formula, the sum of the roots must be 0 , and the fourth root must be -6 . Therefore, the polynomial factors as $(x-3)(x+2)(x-5)(x+6)$. Setting $x=1$ gives $1+a+b+c=(1-3)(1+2)(1-5)(1+6)=168$, so $a+b+c=167$.