Question

What are all the possible rational zeros for `f(x)=x^3-10x^2+16x+15` and how do you find all zeros?

Answer 1

`f(x)` has zeros `x=3` and `x = 7/2+-sqrt(69)/2`

Explanation:

`f(x) = x^3-10x^2+16x+15`

By the rational roots theorem, any rational zeros of `f(x)` are 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.

That means that the only possible rational zeros are:

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

We find:

`f(3) = 27-90+48+15 = 0`

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

`x^3-10x^2+16x+15 = (x-3)(x^2-7x-5)`

We can find the zeros the remaining quadratic `x^2-7x-5` which is in standard form `ax^2+bx+c` with `a=1`, `b=-7` and `c=-5` using the quadratic formula:

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

`color(white)(x) = (7+-sqrt((-7)^2-4(1)(-5)))/(2*1)`

`color(white)(x) = (7+-sqrt(49+20))/2`

`color(white)(x) = 7/2+-sqrt(69)/2`