Suppose f(x) is a quadratic polynomial with leading coefficient 1 such that f(x) + f(-x) = 2x^2 for all x and f(2016) = 0. What is the value of f(10) ?
To find the value of f(10), we can start by using the given information that f(x) + f(-x) = 2x^2 for all x.
Let's substitute x with 10 in the equation f(x) + f(-x) = 2x^2:
f(10) + f(-10) = 2(10)^2
f(10) + f(-10) = 200
Since f(x) is a quadratic polynomial with a leading coefficient of 1, it can be written in the form f(x) = ax^2 + bx + c, where a, b, and c are constants.
Now let's substitute x with 10 and -10 in the quadratic form of f(x):
f(10) = a(10)^2 + b(10) + c
f(-10) = a(-10)^2 + b(-10) + c
Simplifying these equations, we have:
f(10) = 100a + 10b + c
f(-10) = 100a - 10b + c
Now, using the given information that f(2016) = 0, we can substitute x with 2016 in the quadratic form of f(x):
f(2016) = a(2016)^2 + b(2016) + c
0 = 4064256a + 2016b + c
Since the leading coefficient is 1, we can rewrite this equation as:
4064256a + 2016b + c = 0
So now we have a system of three equations:
f(10) + f(-10) = 200
f(10) = 100a + 10b + c
4064256a + 2016b + c = 0
To solve this system of equations, we will use the method of substitution or elimination to find the values of a, b, and c. Once we find the values of a, b, and c, we can substitute them back into the equation f(10) = 100a + 10b + c to find the value of f(10).