Question 1209722
.
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 1/p + 1/q + 1/r + 1/s. 
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~



There is another method to solve, so beautiful that 
the blind begin to walk and the mute begin to see.



<pre>
        Notice that since the constant term is not zero, 
           no one root p, q, r  or  s  is zero.


Reduce the given equation to the standard form combining like terms.  You will get

    g(x) = x^4 + 2x^3 + 16x^2 + 20x - 31.    (1)


Divide this reduced equation by x^4.  You will get another polynomial (like a sock turned inside out)

    {{{1}}} + {{{2/x}}} + {{{16/x^2}}} + {{{20/x^3}}} - {{{31/x^4}}}.    (2)


If some value p, q, r, s  is the root to polynomial (1),
then 1/p, 1/q, 1/r, 1/s  is the zero of function (2).


Let's consider now the polynomial

    h(y) = 1 + 2y + 16y^2 + 20y^3 - 31y^4.    (3)


Compare (3) with (2) and recognize that (3) is the same expression as (2) with replaced  '1/x' by  'y'.


Since p, q, r and s are the roots for polynomial (1),
1/p, 1/q, 1/r and 1/s are the roots for polynomial (3).


Now apply Vieta's theorem and find that the sum of the roots 1/p, 1/q, 1/r, 1/s
is equal to the coefficient 20 at y^3 in polynomial (3), divided by the leading coefficient -31 at y^4,
taken with the opposite sign

    1/p + 1/q + 1/r + 1/s = {{{-20/(-31)}}} = {{{20/31}}}.


At this point, the problem is solved in full, without making cumbersome calculations.


<U>ANSWER</U>.  1/p + 1/q + 1/r + 1/s = {{{20/31}}}.
</pre>

Isn't it beautiful ?



This method is called  "turning a polynomial inside out".


Turning a polynomial inside out can be done mentally, 
so one can write an answer immediately, without making/writing these reasons on paper.


If you show this focus-pocus to your teacher/professor or at the interview,
the other side will be shocked to see such an elegant solution.