Question

What are the zeros in x^3+3x^2-3x-5?

Answer 1

`{-1, -1+sqrt(6), -1-sqrt(6)}`

Explanation:

Applying the rational roots theorem, we find that any rational roots of the given expression will be of the form `p/q` where `p` is a divisor of `-5` and `q` is a positive divisor of `1`. Then, the possible rational roots are `+-1` and `+-5`.

Trying these out, we find that `(-1)^3+3(-1)^2-3(-1)-5 = 0` Thus `x-(-1) = x+1` is a factor of `x^3+3x^2-3x-5`

Dividing, we get
`x^3+3^2-3x-5 = (x + 1)(x^2 + 2x -5)`

To find the remaining roots, we can simply apply the quadratic formula
`ax^2+bx+c = 0 => x = (-b+-sqrt(b^2-4ac))/(2a)`
to obtain the roots of `x^2+2x-5` as

`(-2+-sqrt(2^2-4(1)(-5)))/(2(1)) = -1 +- sqrt(6)`

The the final set of roots (zeros) of the expression is

`{-1, -1+sqrt(6), -1-sqrt(6)}`