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 $(4) + (-5) = -1.$)
A quadratic polynomial has the form: f(x) = ax2 + bx + c
f(-4) = -22 ---> -22 = a(-4)2 + b(-4) + c ---> 16a - 4b + c = -22
f(-1) = 2 ---> 2 = a(-1)2 + b(-1) + c ---> a - b + c = 2
f(2) = -1 ---> -1 = a(2)2 + b(2) + c ---> 4a + 2b + c = -1
Combining the first equation with the third equation:
16a - 4b + c = -22 ---> 16a - 4b + c = -22
4a + 2b + c = -1 ---> x -4 ---> -16a - 8b - 4c = 4
Adding down the columns: -12b - 3c = -18 ---> 4b + c = 6
Combing the second equation with the third equation:
a - b + c = -2 ---> x -4 ---> -4a + 4b - 4c = -8
4a + 2b + c = -1 ---> 4a + 2b + c = -1
Adding down the columns: 6b - 3c = -9 ---> 2b - c = -3
Combining these two new equations: 4b + c = 6
2b - c = -3
Adding down the columns: 6b = 3 ---> b = 0.5
Substituting back: c = 4 and a = -1.5
You'll need to find the sum.
A quadratic polynomial has the form: f(x) = ax2 + bx + c
f(-4) = -22 ---> -22 = a(-4)2 + b(-4) + c ---> 16a - 4b + c = -22
f(-1) = 2 ---> 2 = a(-1)2 + b(-1) + c ---> a - b + c = 2
f(2) = -1 ---> -1 = a(2)2 + b(2) + c ---> 4a + 2b + c = -1
Combining the first equation with the third equation:
16a - 4b + c = -22 ---> 16a - 4b + c = -22
4a + 2b + c = -1 ---> x -4 ---> -16a - 8b - 4c = 4
Adding down the columns: -12b - 3c = -18 ---> 4b + c = 6
Combing the second equation with the third equation:
a - b + c = -2 ---> x -4 ---> -4a + 4b - 4c = -8
4a + 2b + c = -1 ---> 4a + 2b + c = -1
Adding down the columns: 6b - 3c = -9 ---> 2b - c = -3
Combining these two new equations: 4b + c = 6
2b - c = -3
Adding down the columns: 6b = 3 ---> b = 0.5
Substituting back: c = 4 and a = -1.5
You'll need to find the sum.