find all zeros of f(x)=x^4-1
You need to solve the equation `x^4 - 1 = 0` , to find the zeroes of function.
You should notice that the equation of function is a difference of squares, hence you may write the equation using the formula `u^2 - v^2 = (u-v)(u+v).` Hence, `x^4 - 1 = (x^2-1)(x^2+1)`
You need to solve the equation `(x^2-1)(x^2+1) = 0`
You should notice that the equation `(x^2-1)(x^2+1)` may be factorized using the formula of difference of squares such that:
`x^2-1 = (x-1)(x+1)`
You need to solve the equation `(x-1)(x+1)(x^2+1) = 0`
`x-1 = 0 =gt x_1 = 1`
`x+1 = 0 =gt x_2 = -1`
`x^2 + 1 = 0 =gt x^2 = -1 =gt x_(3,4) = +-sqrt(-1)`
Using the complex number theory yields `x_(3,4) = +-i.`
Sketching the graph of the function, you may see where the graph of function intercepts x axis : `x_1=1 and x_2=-1` .
Hence, evaluating the zeroes of the equation of function yields `x_1 = 1; x_2 = -1; x_(3,4) = +-i.`