Question 1165974
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Let f(x) be a quadratic polynomial such that f(-4) = -22, f(-1)=2, and f(2)=-1. Let g(x) = f(x)^{16}. 
Find the sum of the coefficients of the terms in g(x) with even exponents. 

(For example, the sum of the coefficients of the terms in -7x^3 + 4x^2 + 10x - 5 with even exponents is (4) + (-5) = -1.)
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The steps are as follows:


<pre>
    1)  First we find the quadratic polynomial  f(x) via its coefficients  f(x) = ax^2 + bx + c.


    2)  Then we find the value of f(1),   which is the sum of its coefficients f(1) = a + b + c.


    3)  Next, we find the value of f(-1), which is an ALTERNATE sum of its coefficients  f(-1) = a - b + c.


    4)  Finally, we will find [f(1)]^16 = (a + b + c)^16  and  [f(-1)]^16 = (a - b + c)^16.


        Then we note that the sum of the coefficients of the terms in g(x) with even exponents is exactly half of the sum  [f(1)]^16 + [f(-1)]^16,
        since the odd exponent coefficients will cancel each other when summing.
</pre>


Now I will implement these steps.



<pre>
Let f(x) = ax^2 + bx + c.


First, we want to find coefficients "a", "b" and "c".


We have  these equations:


    at x= -4:  16a - 4b + c = -22

    at x= -1:    a -  b + c =   2

    at x=  2:   4a + 2b + c =  -1


You can solve the system by any method you want. You will get then   a = -3/2,  b= 1/2,  c= 4.



Hence, the value of f(1) is  {{{-3/2 + 1/2 + 4}}} = 3.


       The value   f(-1) is just given by the condition  f(-1) = 2.


It implies that the sum  [f(1)]^16 + [f(-1)]^16  is  {{{3^16}}} + {{{2^16}}}.


Hence, the answer to the problem's question is  {{{(1/2)*(3^16 + 2^16)}}} = {{{43112257/2}}}..


<U>ANSWER</U>.  The sum of the coefficients of the terms in g(x) with even exponents is  {{{(1/2)*(3^16 + 2^16)}}} = {{{43112257/2}}}.
</pre>


Solved.



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Usually, &nbsp;such problems are considered as &nbsp;"coffin" &nbsp;problems, &nbsp;meaning that the average school student does not know how to solve them.


Only those are able to solve, &nbsp;who know this trick with alternate sums of the coefficients of polynomials.


Those who study &nbsp;Math from &nbsp;Math circles, &nbsp;know it . . .