Question

How do you find all the zeros of `f(x) = 4(x + 3)^2 - 1`?

Answer 1

Set `f(x) = 0` and solve for `x` to find that `f(x)` has zeroes at `x=-7/2` and `x=-5/2`

Explanation:

`4(x+3)^2-1 = 0`

`=> 4(x+3)^2 = 1`

`=> (x+3)^2 = 1/4`

`=> x+3 = +-sqrt(1/4) = +-1/2`

`=>x = -3+-1/2`

`:. x = -7/2` or `x = -5/2`