Question

How do you find the zeros of `f(x) = 2x^3 − 5x^2 − 28x + 15`?

Answer 1

Zeros: `x=1/2`, `x=5`, `x=-3`

Explanation:

`f(x) = 2x^3-5x^2-28x+15`

By the rational root theorem, since `f(x)` has integer coefficients, any rational zeros 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 `2` of the leading term.

That means that the only possible rational zeros of `f(x)` are:

`+-1/2, +-1, +-3/2, +-5/2, +-3, +-5, +-15/2, +-15`

Trying the first one we find:

`f(1/2) = 2/8-5/4-28/2+15 = (1-5-56+60)/4 = 0`

So `x=1/2` is a zero and `(2x-1)` a factor:

`2x^3-5x^2-28x+15`

`=(2x-1)(x^2-2x-15)`

To factor the remaining quadratic, find a pair of factors of `15` which differ by `2`. The pair `5, 3` works, so:

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

So the two remaining zeros are `x=5` and `x=-3`