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

2 Answers
Mar 20, 2017

Answer:

#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

Answer:

#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)#.