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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 $(4) + (-5) = -1.$)

 Jun 18, 2020
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The quadratic polynomial f(x) is in the form \(ax^2+bx+c\). We can derive an equation from each output in the form \(a^2+b+c\). The equation for \(f(-4)=-22\) is \(16a-4b+c=-22\). The equation for f(-1)=2 is a-b+c=2. The equation for f(2)=-22 is 4a+2b+c=-1.

 

We have a system of equations now so we solve for a, b, and c.

 

 \(16a-4b+c=-22\\a-b+c=2\\4a+2b+c=-1\)

 

We find that \(a= -\frac{3}{2}, b=\frac{1}{2}, \text{and}\ c=4\) . Thus the equation that represents this function is \(f(x)=-\frac{3}{2}x^2+\frac{1}{2}x+4\). We know that g(1) gives us the sum of all coefficients of the terms in g(x). To get the sum of the coefficients of the terms in g(1) with an even degree, we must add g(-1). \(g(x)=f(x)^{16} \text{ so}\ g(x)=(-\frac{3}{2}x^2+\frac{1}{2}x+4)^{16}\). To get only the terms with even degrees, we add g(-1) to g(1). g(-1) is

 

 \(g(x)=f(x)^{16}\\g(-1)=f(-1)^{16}\\g(-1)=2^{16}.\)

 

g(1) is

 \(g(x)=f(x)^{16}\\g(1)=f(1)^{16}\\g(1)=(-\frac{3}{2}x^2+\frac{1}{2}x+4)^{16}\\g(1)=(-\frac{3}{2}(1)^2+\frac{1}{2}(1)+4)^{16}\\g(1)=3^{16}.\)

 

We add g(-1), which is 2^{16}, to g(1), which is 3^{16}. However, we need to divide this by 2 because when we added g(-1) to g(1), this gave us double of the sum of the coefficients of the terms with even degree. Therefore the answer is \(\boxed{\frac{3^{16}+2^{16}}{2}}\).

 Jun 18, 2020

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