Well, we see square roots, so let's square our expression for : By Vieta's, we have , so the expression simplifies to

Hrm. We don't know what the value of is. It's certainly not , and there's no way we can find the polynomial whose roots are , , and , cause that would be an degree polynomial. Not good.

Let's step away from and focus on the polynomial we're trying to find: . We can factor an to get , and then we can use Rational Root Theorem to factor even further and get . However, it's not clear what we're going to do now. We might have to go back to and find what is.

But wait. We don't need to evaluate it directly. We just need to find it. Whenever we've encountered expressions we didn't know before, we've set variables equal to them and solved. Let's do that here. Let . This means that By Vieta's, we have and , so the expression simplifies to .

Bingo! The expression inside the parenthesis is just ! Now we can set up a system! In particular, we can take the two big expressions we have and substitute all occurrences of and to get the system Solving this gives us... uh oh. There's no neat solution. By neither addition nor subtraction do we get a clean equation. Nor can we square either equation, since that still leaves a degree term that we don't know what to do with. Additionally, taking the square root of both sides won't work, since then we'd have to do casework (positive square root or negative square root?) and that becomes messy and tedious. But as alluded to earlier, the polynomial we're trying to evaluate is a little weird... ah we have an expression for , let's plug it in.