21st AIME problem: roots of x^4 - x^3 - x^2 - 1 = 0 are....

malick

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9) The roots of x<sup>4</sup> - x<sup>3</sup> - x<sup>2</sup> - 1 = 0 are a, b, c, and d. Find p(a) + p(b) + p(c) + p(d), where p(x) = x<sup>6</sup> - x<sup>5</sup> - x<sup>3</sup> - x<sup>2</sup> - x.
This was a past AIME problem. An answer was as follows:

. . . . .x<sup>4</sup> - x<sup>3</sup> - x<sup>2</sup> - 1 = (x + 1)(x<sup>3</sup> - 2x<sup>2</sup> + x - 1), so:

. . . . . . .a = -1
. . . . . . .b + c + d = 2
. . . . . . .bc + cd + db = 1

. . . . .Hence b<sup>2</sup>+c<sup>2</sup>+d<sup>2</sup> = 2<sup>2</sup> - 2 = 2.

. . . . .We have:

. . . . . . .p(x) = (x<sup>3</sup> - 2x<sup>2</sup> + x - 1)(x<sup>3</sup> + x<sup>2</sup> + x + 1) + x<sup>2</sup> - x + 1

. . . . .Hence:

. . . . . . .p(a) = 3
. . . . . . .p(b) = b<sup>2</sup> - b + 1
. . . . . . .p(c) = c<sup>2</sup> - c + 1
. . . . . . .p(d) = d<sup>2</sup> - d + 1

. . . . .so:

. . . . .p(a) + p(b) + p(c) + p(d)
. . . . . . .= 6 + (b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>) - (b + c + d)
. . . . . . .= 6

Can someone please esplicitly explain this to me.
 
It might help if you said where you're bogging down.

For instance, you can do the factoring, so the first step is okay, right? But are you not familiar with the formulas for "given these roots, those coefficients have to be in that form"? (This problem expects you to have those formulae memorized, but many students have never seen them, is why I ask.)

Thank you.

Eliz.
 
This may help you to understand what is going on.
\(\displaystyle \L
\begin{array}{rcl}
(x - a)(x - b)(x - c)(x - d) & = & x^4 \\
& + & ( - a - b - c - d)x^3 \\
& + & (ab + ac + ad + bc + bd + cd)x^2 \\
& + & ( - abc - abd - acd - bcd)x \\
& + & (abcd) \\
\end{array}\).
Now the coefficients of the polynomials are equal.
 
then again I still don't get the question completely. What does this mean, and how can you use it to solve the problem.
 
malick said:
What does this mean...?
What does which mean? How far have you gotten? Where are you stuck?

Please be specific. Thank you.

Eliz.
 
I can do the factoring, so the first step is fine. Like you said, I am not familiar with the formulas and how to get the other roots. How would you do this problem?
 
malick said:
How would you do this problem?
You would do it the way the solution showed. However, it appears that, to be understood, the solution requires quite a lot of material that you haven't yet studied, such as rules regarding roots, how to multiply polynomials, how to equate coefficients, and how to solve systems of equations.

I'm afraid it is beyond our abilities to teach you (in this environment) the various courses required to cover this material. You might want to consider hiring a tutor, local to your area. Then you can spend a few hours a week (or every day), and catch up on all of the necessary background material.

Eliz.
 
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