Question

How do you solve `f(n)=-3n^4-6n^3-3n^2-9n-7` for `n=-2`?

Answer 1

`f(-2) = 5`

Explanation:

For each `n` in the equation substitute `-2` to give:

`f(-2) = -3(-2^4) - 6(-2^3) - 3(-2^2) - 9(-2) - 7`

`f(-2) = (-3(-2*-2*-2*-2)) - (6(-2*-2*-2)) - (3(-2*-2)) - (9*-2) - 7`

`f(-2) = (-3*16) - (6*-8) - (3*2) + 18 - 7`

`f(-2) = -48 + 48 - 6 + 18 - 7`

`f(-2) = 5`