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.$)

The sum of the coefficients of the terms in (-1.5·x² + 0.5·x + 4) ¹⁶ that have even degree is - 4273411167.501

Step-by-step explanation:

The parameters given are;

f (-4) = - 22

f (-1) = 2

f (2) = - 1

g (x) = f (x) ¹⁶

The function f (x) is presented as follows;

f (x) = a·x² + b·x + c

We have;

-22 = a· (-4) ² + b· (-4) + c

-22 = a·16 - 4·b + c ... (1)

2 = a· (-1) ² + b· (-1) + c

2 = a - b + c ... (2)

-1 = a· (2) ² + b· (2) + c

-1 = 4·a + 2·b + c ... (3)

Solving the equations (1), (2), and (3) by using an online linear systems solver, we get;

a = - 1.5, b = 0.5, c = 4

Therefore, f (x) = - 1.5·x² + 0.5·x + 4

f (x) ¹⁶ = (-1.5·x² + 0.5·x + 4) ¹⁶ which gives the coefficients of the even terms as follows;

656.841 - 19267.331 + 248302.054 - 1772904.419 + 6735603.932 - 2868054.635 - 119602865.901 + 750783340.827 + - 2542435585.611 + 5338903756.992 - 6048065910.25 - 1031335136 + 17223697920 - 32238338048 + 32107397120 - 17716740096 = - 4273411167.501.