Question

What are all the zeroes of `g(x)= 2x^3 - 5x^2 + 4`?

Answer 1

`g(x)` has zeros `2` and `1/4+-sqrt(17)/4`

Explanation:

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

By the rational root theorem, any rational zeros of `g(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `4` and `q` a divisor of the coefficient `2` of the leading term.

That means the only possible rational zeros are:

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

We find:

`g(2) = 2(8)-5(4)+4 = 16-20+4 = 0`

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

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

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

`Delta = b^2-4ac = (-1)^2-4(2)(-2) = 1+16 = 17`

Since this is positive but not a perfect square, the remaining zeros are Real but irrational. They are given by the quadratic formula:

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

`= (1+-sqrt(Delta))/4`

`= (1+-sqrt(17))/4`

`= 1/4+-sqrt(17)/4`