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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) that have even degree. (For example, the sum of the coefficients of the terms in -7x^3 + 4x^2 + 10x - 5 that have even degree is (4) + (-5) = -1.)

 

My findings : f(x) = -3/2x^2 + 1/2x + 4. 

 Jun 1, 2020
 #1
avatar+30903 
+1

You have the correct f(x).

 

To find the sum of the even coefficients of f(x)^16 simply calculate (f(1)^16  +  f(-1)^16)/2.

 Jun 1, 2020
 #2
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Thanks Alan! Why would we need to divide by 2?

 Jun 1, 2020
 #3
avatar+30903 
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Because for both f(1)^16 and f(-1)^16 the even power terms will have the same sign, the odd power terms will have the opposite sign. When you add the two together the odd terms will cancel, but the even terms will double. Hence you need to divide by 2.

Alan  Jun 1, 2020
 #5
avatar+30903 
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Yes.

Alan  Jun 1, 2020
 #4
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So would the answer be 43112257/2?

 Jun 1, 2020

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