Question

How do you find the zeros of `x^3-x^2-7x+15=0`?

Answer 1

`x=-3` and `x=2+-i`

Explanation:

`f(x) = x^3-x^2-7x+15`

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p`, `q` with `p` a divisor of the constant term `15` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational zeros are:

`+-1`, `+-3`, `+-5`, `+-15`

Trying each of these in turn, we find:

`f(-3) = -27-9+21+15 = 0`

So `x=-3` is a zero and `(x+3)` a factor:

`x^3-x^2-7x+15 = (x+3)(x^2-4x+5)`

We can factor `x^2-4x+5` by completing the square and using the difference of squares identity:

`a^2-b^2=(a-b)(a+b)`

with `a=(x-2)` and `b=i` as follows:

`x^2-4x+5 = x^2-4x+4+1 = (x-2)^2-i^2 = (x-2-i)(x-2+i)`

Hence zeros:

`x = 2+-i`