Polynomial Question sum of powers 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 $(4) + (-5) = -1.$)
I have determined the quadratic expression, which is $$f(x)=-\frac{3x^2}{2}+\frac{x}{2}+4.$$but I don't know how to proceed.
EDIT: What I am learning is related to odd and even polynomials. Is there a way to solve this problem by reducing $g(x)$ to an even polynomial?
 A: The key is the following observations:

The sum of all coefficients of a (one variable) polynomial, is equal to that polynomial evaluated at $1$.

and

The difference between the coefficients of powers of $x$ with even exponents and powers of $x$ with odd exponents, is equal to that polynomial evaluated at $-1$.

Why? Suppose that $g(x) = a_nx^n + ... + a_{0}$, then $g(1) = a_n + a_{n-1}  + ... + a_0$, the sum of all coefficients. Similarly, $g(-1) = \sum_{k\  \mathrm{even}}^{k \leq n} a_k - \sum_{k\  \mathrm{odd}}^{k \leq n} a_k$, simply by noting what the powers of $1$ and $-1$ are, respectively.
Therefore, $g(1) = \sum_{k=0}^n a_k = \sum_{k\  \mathrm{even}}^{k \leq n} a_k + \sum_{k\  \mathrm{odd}}^{k \leq n} a_k$. It follows that $\sum_{k\  \mathrm{even}}^{k \leq n} a_k = \frac{g(1)+ g(-1)}{2}$.
You know that $g(x) = f(x)^{16}$, and you know what $f(x)$ is, so finish.
A: First of all, we need to determine what quadratic $f(x)$ is. We can write some equations in the form of a general quadratic($ax^2+bx+c$). We plug in the values stated in the question into our expression to get the following system of linear equations:

*

*$16a-4b+c=-22.$

*$a-b+c=2.$

*$4a+2b+c=-1.$
Now, we can solve this set of equations and find out what the coefficients of our expression is. We find that:
$$a=-\frac{3}{2}, b=\frac{1}{2}, c=4.$$So:
$$f(x)=-\frac{3x^2}{2}+\frac{x}{2}+4.$$
Now that we have our quadratic expression, we can look at $g(x)$. Since we only want the sum of the even exponents, we can try the values $f(1)$ and $f(-1)$. The first value works in all cases, while the negative value will resulting in the sum of the odd exponents subtrated from the even exponents.
We substitute $x=1$ into our quadratic to get $f(1)=3$, so:
$$(f(1))^{16}=3^{16}.$$
Now, we substitute $x=-1$, to get $f(-1)=2$, so:
$$(f(-1))^{16}=2^{16}.$$
Now that we have our two equations, we know that $f(1)-f(-1)=3^{16}-2^{16}$, which is also two times the sum of the coefficients of the odd terms. So the sum of the coefficients of the odd terms is:
$$\frac{3^{16}-2^{16}}{2}.$$
We subtract this from $f(1)$ and simplify to get the sum of the even exponents.
$$3^{16}-\frac{3^{16}-2^{16}}{2}=\frac{3^{16}+2^{16}}{2}.$$
Thus, the sum of the coefficients of the even exponents in $g(x)$ is $\boxed{\frac{3^{16}+2^{16}}{2}}$*.
*I used a calculator to find the result in numerical form: $\boxed{21556128.5}$.
A: Using this you find that the answer is
$\frac{1}{2} (f(1)^{16}+f(-1)^{16})=\frac{1}{2} (2^{16}+3^{16})$
