Question 1209724
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Let p, q, r, and s be the roots of g(x) = 3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14.
Compute p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2. 
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<pre>
Reduce the polynomial to the standard form, combining like terms

    3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14 = x^4 + 2x^3 + 16x^2 + 20x - 31.


Notice that 

    p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2 = pqrs*(p + q + r + s).


Also notice that the leading coefficient of the polynomial standard form at x^4 is 1.


Due to Vieta's theorem

    pqrs          = -31  (the product of the roots is equal to the constant term)

    p + q + r + s =  -2  (the sum of the roots is the coefficient at x^3 with the opposite sign).


Therefore,

    p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2 = (-31)*(-2) = 62.


<U>ANSWER</U>.  p^2*qrs + pq^2*rs + pqr^2*s + pqrs^2 = 62.
</pre>

At this point, the problem is solved completely.



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Edwin, I explain it to you, again.


The original problem was in this reduced polynomial form.


They composed this their monstrous version in order for to create their OWN NEW problem for their web-site.



- .-.- .-.- .-.- .-.- .-.- .-.- .-.- .-.- .-.- .-.



Edwin, you ask me HOW do I know.


First, it is obvious even for hedgehog or for rabbit.


Second, I have an ability to see through walls.



Really, I can see what happens on the other side of a wall.

I can do it not because the rays of light go through a wall.

I can see, because I know what happens on the other side of a wall.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(using the method of deduction . . . )



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Knowledge is a power.