Question

How do you find all the zeros of `2x^3+9x^2+6x-8`?

Answer 1

This cubic has zeros `-2` and `1/4(-5+-sqrt(57))`

Explanation:

`f(x) = 2x^3+9x^2+6x-8`

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 `-8` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2, +-1, +-2, +-4, +-8`

We find:

`f(-2) = 2(-8)+9(4)+6(-2)-8 = -16+36-12-8 = 0`

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

`2x^3+9x^2+6x-8=(x+2)(2x^2+5x-4)`

The remaining quadratic is in the form `ax^2+bx+c` with `a=2`, `b=5`, `c=-4`.

We can solve this using the quadratic formula:

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

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

`=1/4(-5+-sqrt(25+32))`

`=1/4(-5+-sqrt(57))`