Reposted again bc a certain user kept complaining about my formatting (btw it's fixed now, stop responding to my posts jfan17.)

I'm responding to your posts because we are not here to fix your lousy formatting, we're here to help you with the problems. If you want me to continue helping you out, I'd suggest you to not copy paste directly from whatever site you're using and cheating off of.

Name \(f(x) = ax^2 + bx + c\)

\(f(-4) = -22 \)

\(f(-1) = 2\)

\(f(2) = -1\)

Substituting -4 for x in our first equation, we get:

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

Substituting -1 for x in our second equation, we get:

\(a-b+c = 2\)

Finally, substituting 2 for our third equation, we get:

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

First we can subtract equations (3) and (2) to get:

\(3a +3b = -3\)

Dividing by 3 on both sides, we then get:

\(a+b = -1\)

\(a = -1 -b\)

Substitute into our first equation, we get:

\(-16-16b-4b+c = -22\)

\(-16-20b + c = -22\)

Substitute back into our second equation, we get:

\(-1-2b + c = 2\)

Subtracting our first equation from our second equation, we then arrive at:

\(15+18b =24\)

\(18b = 9 \)

\(b = \frac1{2}\)

Now that we have b, we're looking for a and c.

If we substitute back into our equation

\(a+b = -1\)

\(a + \frac1{2} = -1\)

\(a = -\frac3{2}\)

Substitute into any of our equations(I'll use the second equation for simplicity)

\(-\frac3{2} -\frac1{2} + c = 2\)

\(-2 + c = 2 \)

\(c = 4\)

We have the quadratic:

\(f(x) = -\frac3{2}x^2 + \frac1{2}x + 4\)

We need to take the 16th power of this quadratic. 

g(x) = f(x)¹⁶

We can write :

g(x) = a₀ + a₁x + a₂x² +..........

If we plug in 1 into this equation, we get:

g(1) = a₀ + a₁ + a₂ +..........

3¹⁶ = a₀ + a₁ + a₂ +..........

If we plug in -1 into the function g(x), we get:

g(-1) = a₀ - a₁ + a₂ +..........

2¹⁶ = a₀ - a₁ + a₂ +..........

If we add these two equations, we see that all of the odd coefficients cancel out, leaving us with only the even ones. We then get,

43112257 = 2(a₀ + a₂ +..........)

Our answer is then 43112257/2.