Question

How do you find all zeros of `f(x)=3x^3-12x^2+3x`?

Answer 1

`x=2.535`

`x=.131`

Explanation:

To find the zeros of a function, we must first find the derivative of the function. This one is not very difficult, we will just apply the power rule to each individual term

`f(x)=3x^3 - 12x^2 + 3x`
`f'(x)=9x^2 - 24x +3`

Then we set the derivative equal to zero and solve for x

`9x^2 - 24x + 3 = 0`

We can plug this into our calculator to find the two x values seeing as how this function is not factorable. We get two values

`x=2.535` `x=.131`

Answer 2

`x=0, x=2+sqrt(3), and x=2-sqrt(3)`

Explanation:

First, factor out an x:
`f(x)=x(3x^2-12x+3x)`
Already, you can figure out that when x zero, f(x) is zero, so `x=0` is one solution.
Then, attempt to factor the quadratic. Unfortunately, in this case, you can't use factoring, so plug the numbers into the quadratic formula, which is `x=(-b+-sqrt(b^2-4ac))/2a` in case you don't have it memorized:
`x=(-(-12)+-sqrt((-12)^2-4(3)(3)))/(2(3))`
`x=(12+-sqrt(144-36))/(2(3))`
`x=(12+-sqrt(108))/(6)`
(The square root can be simplified to `6sqrt(3)`, because `sqrt(108)=sqrt(6*6*3)=sqrt(6)^2*sqrt(3)=6*sqrt(3)`)
`x=(12+-6sqrt(3))/(6)`
A six can also be cancelled out:
`x=(2+-sqrt(3))`
That means the zeroes are `x=0, x=2+sqrt(3), and x=2-sqrt(3)`.