Finding the sum of squares of roots of a quartic polynomial.

What is the sum of the squares of the roots of $ x^4 - 8x^3 + 16x^2 - 11x + 5 $ ?

This question is from the 2nd qualifying round of last year's Who Wants to be a Mathematician high school competition which can be seen here:

I know the answer (32) because that is also given in the link, and I have checked by brute force that the given answer is correct.

However, I have made no progress at all in figuring out how to calculate the sum of squares of the roots - either before or after knowing the answer! I was expecting there to be a nice "trick" analagous to the situation if they had given a quadratic and asked the same question -- in that case I know how to get the sum and product of the roots directly from the coefficients, and then a simple bit of algebraic manipulation to arrive at the sum of squares of the roots.

In this case (the quartic) I have no idea how to approach it, and I have not spotted any way to simplify the problem (e.g. I cannot see an obvious factorisation, which might have helped me).

I've looked on the web at various articles which dicuss the relationships between the coefficients of polynomials and their roots and - simply put - I found nothing which gave me inspiration for this puzzle.

Given the audience for this test, it should be susceptible to elementary methods ... I would appreciate any hints and/or solutions!

Thank you.

Answer

We have that

$$(x-a)(x-b)(x-c)(x-d)=$$
$$=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2\\-(abc+abd+acd+bcd)x+abcd$$

then by

• $S_1=a+b+c+d$


• $S_2=ab+ac+ad+bc+bd+cd$


• $S_3=abc+abd+acd+bcd$


• $S_4=abcd$

$$a^2+b^2+c^2+d^2=S_1^2-2S_2$$

and more in general by Newton's sums we have that

• $P_1=a+b+c+d=S_1$


• $P_2=a^2+b^2+c^2+d^2=S_1P_1-2S_2$


• $P_3=a^3+b^3+c^3+d^3=S_1P_2-S_3P_1+3S_3$


• $P_4=a^4+b^4+c^4+d^4=S_1P_3-S_2P_2+S_3P_1-4S_4$