Question

How do you find all the zeros of `x^3+2x^2-2x-3`?

Answer 1

Use the rational root theorem to help find the first zero `x=-1`, then divide `f(x)` by `(x+1)` to find a quadratic and hence the other two zeros:

`x = (-1+-sqrt(13))/2`

Explanation:

`f(x) = x^3+2x^2-2x-3`

By the rational root theorem, any rational zeros of `f(x)` will be expressible in the form `p/q` for some integers `p` and `q` with `p` a divisor of the constant term `-3` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`+-1`, `+-3`

Trying each in turn, we find:

`f(1) = 1+2-2-3 = -2`

`f(-1) = -1+2+2-3 = 0`

So `x=-1` is a zero and `(x+1)` a factor:

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

The remaining quadratic factor is of the form `ax^2+bx+c` with `a=1`, `b=1` and `c = -3`. This has zeros given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

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

`=(-1+-sqrt(13))/2`