Question

How do you find all the zeros of `f(x) = x^3 – 12x^2 + 28x – 9`?

Answer 1

Use the rational root theorem to help find the zero `x=9`, then factor to find the other zeros: `x=3/2+-sqrt(5)/2`

Explanation:

`f(x) = x^3-12x^2+28x-9`

By the rational root 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 `-9` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible rational zeros are:

`+-1`, `+-3`, `+-9`

In addition, note that:

`f(-x) = -x^3-12x^2-23x-9`

has all negative coefficients. So `f(x)` has no negative zeros.

So the only possible rational zeros are:

`1, 3, 9`

Trying each in turn, we find:

`f(9) = 729-972+252-9 = 0`

So `x=9` is a zero and `(x-9)` a factor of `f(x)`:

`x^3-12x^2+28x-9 = (x-9)(x^2-3x+1)`

We can find the remaining two zeros using the quadratic formula:

`x = (3+-sqrt((-3)^2-4(1)(1)))/(2*1)`

`=(3+-sqrt(9-4))/2`

`=3/2+-sqrt(5)/2`