What is the sum of polynomial zeros?

[Amad27]

There are nonzero integers $a$, $b$, $r$, and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x) = x^3 - ax^2 + bx - 65$. For each possible combination of $a$ and $b$, let $p_{a,b}$ be the sum of the zeroes of $P(x)$. Find the sum of the $p_{a,b}$'s for all possible combinations of $a$ and $b$.

From Vieta's Formulas, I got:

$a=2r+k$
$b=2rk+r^2+s^2$
$65=k(r^2+s^2)$

Where $k$ is the other real zero.

Then I split it into several cases: $r^2 + s^2 = 1, 5, 13, 65$ then:

For case 1: $r = \{2, -2, 1, -1 \}$

$\sum a = 2(\sum r) + k \implies a = 13$

Then for case 2: $r^2 + s^2 = 13$, it is that,

$\sum a = 2(\sum r) + k \implies a = 5$

Case 3:

$\sum a = 2(\sum r) + k \implies a = 65$

Case 4:

$\sum a = 2(\sum r) + k \implies a = 1$

$$\sum a = 1 + 65 + 5 + 13 = 84$$

The correct answer is $80$, why?

 

[anemone]

Hi Olok,

Yes, the problem actually asked us to find the value $\sum a$.

You're definitely in the right track when you considered different cases for $r^2 + s^2$ and $k$ since we got $k(r^2+s^2)=65=1(65)=(65)(1)=5(13)=13(5)$.

But you must bear in mind that:

1. The question told us $a,\,b,\,r,\,s$ are nonzero integers, here, we could cross out the possibility that $k(r^2+s^2)=(65)(1)$ since $r^2+s^2\ne 1$.

2. When we're dealing with the squares, such as in this instance and take for example when we have $k(r^2+s^2)=(5)(13)=5(2^2+3^2)=5((-2)^2+(-3)^2)$, the target expression $a=2r+k$ is equivalent to $2k$ since the sum of both the values of $r$ equals zero:

$a=2r+k=2(2)+5+2(-2)+5=2(5)$

Therefore, we only need to consider all the possible positive values for $r$ to determine the corresponding $k$ and then multiply $k$ by 2 to get the answer.

And note that we have:

$k(r^2+s^2)=(1)(65)=1(1^2+8^2)=1(8^2+1^2)=1(4^2+7^2)=1(7^2+4^2)$

$k(r^2+s^2)=(5)(13)=5(2^2+3^2)=5(3^2+2^2)$

$k(r^2+s^2)=(13)(5)=13(1^2+2^2)=5(2^2+1^2)$

Therefore,

$\sum a = 2(4(1)+2(5)+2(13))=2(40)=80$

Edit: This thread is previously posted in the Pre-Calculus forum when it should be in the Pre-Algebra and Algebra forum, I have since moved it here, and we hope the OP will take care of which the forum is best suit for the topic in the future as this makes MHB a more organized and thus a more useful place for everyone. Thank you.