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

2 Answers
Mar 20, 2017

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

Mar 20, 2017

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).