SOLUTION: 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. (Fo

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 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. (Fo      Log On


   

Question 1165974: 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.)
Please explain in detail
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52810) About Me  (Show Source):
You can put this solution on YOUR website!
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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: 

    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.


Now I will implement these steps. 


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%2F2+%2B+1%2F2+%2B+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%5E16 + 2%5E16.


Hence, the answer to the problem's question is  %281%2F2%29%2A%283%5E16+%2B+2%5E16%29 = 43112257%2F2..


ANSWER.  The sum of the coefficients of the terms in g(x) with even exponents is  %281%2F2%29%2A%283%5E16+%2B+2%5E16%29 = 43112257%2F2.


Solved.


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

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

Those who study  Math from  Math circles,  know it . . .



Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


Terrific problem. I learned some new mathematics thinking about it and working on it.

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NOTE: The method I used below to solve the problem is of my own design. It seems to me it should be valid.

I will be happy to see if another tutor shows a solution that is different than mine and perhaps shows my answer to be wrong.

I used an online tool to solve the problem, and its answer was VERY CLOSE to mine. Perhaps that tool did some rounding in its calculations... or perhaps my answer was close only by chance....

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We first need to find the quadratic function f(x) defined by the three given points.

f%28x%29+=+ax%5E2%2Bbx%2Bc

f(-4): 16a-4b%2Bc+=+-22
f(-1): a-b%2Bc+=+2
f(2): 4a%2B2b%2Bc+=+-1

Since you are working on a problem like this, I will assume you know how to solve that system of equations, so I won't go through the details. The function is

f%28x%29+=+%28-3%2F2%29x%5E2%2B%281%2F2%29x%2B4

Now we need to find the function

g%28x%29+=+%28f%28x%29%29%5E16

and find the sum of the coefficients of the EVEN degree terms.

This was a new problem to me. Finding the sum of ALL the coefficients is easy -- substitute 1 for each variable and evaluate the expression. So the sum of ALL the coefficients of g(x) would be

%28%28-3%2F2%29%2B%281%2F2%29%2B4%29%5E16+=+3%5E16+=+43046721

But we need to find the sum of only the coefficients of the even degree terms. How are we going to do that?

Then the inspiration hit.

If we change the sign of the linear term in the polynomial, we will get the same expansion except that all the terms of odd degree will have opposite signs. So the sum

[1]

will have the coefficients of the even degree terms doubled and the coefficients of the odd degree terms canceled out.

So the sum of the coefficients of the even degree terms will be half the sum of the coefficients of the expression [1].

And we can evaluate that sum by substituting 1 for each variable everywhere.



And then, finally, the sum of the coefficients of the even degree terms of g(x) is

43112257%2F2

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Many thanks to tutor @ikleyn for showing an easier path to the same answer I got!