Help finding zeros of f(x) = 3x^3 - 12x^2 + 3x

[jwpaine]

In class we are finding the zeros of functions: we did the first couple in class, example:

h(x) = 2x^4-2x^2-40
so we can pull a 2 out of it to get 2(x^4 - x^2 - 20)
and then into a binomial: 2(x^2+4)(x^2-5)
So the zeros of the function are x^2 = -4 which we can forget about... and x^2 = sqrt(5) which works

now I need help doing the same with this polynomial function:

f(x) = 3x^3 - 12x^2 + 3x

I pulled a 3 out of it to give me 0= 3(x^3 - 4x^2 + x)

Where do I go from here?

 

[skeeter]

Re: Help finding zeros of a third degree polynomial

In class we are finding the zeros of functions: we did the first couple in class, example:

h(x) = 2x^4-2x^2-40
so we can pull a 2 out of it to get 2(x^4 - x^2 - 20)
and then into a binomial: 2(x^2+4)(x^2-5)
So the zeros of the function are x^2 = -4 which we can forget about... and x^2 = sqrt(5) which works

now I need help doing the same with this polynomial function:

f(x) = 3x^3 - 12x^2 + 3x

I pulled a 3 out of it to give me 0= 3(x^3 - 4x^2 + x)

Where do I go from here?

try factoring out 3x instead of just 3

 

[jwpaine]

Thanks.

Ok.. so I factored out a 3x

3x(x^2 - 4x +3)

and then factored my trinomial into two binomials:

3x(x-3)(x-1)

so I assume the zeros of this function, are x = {3,1} ???

 

[skeeter]

you missed a solution ... there are three.

 

[jwpaine]

Yes, of course... according to the fundamental theorem of algebra... there are exactly as many complex roots as its degree... OK... I should have 3 solutions.....

The zeros: 3 and 0 make f(3) and f(0) = 0

but f(1) doesn't = 0 for some reason... even though I factored a (x-1)

 

[skeeter]

you factored incorrectly ...

3x³ - 12x² + 3x = 0

3x(x² - 4x + 1) = 0

so ... try again.

 

[jwpaine]

hehe... oops:

I was rushing..... (x^2 - 4x +1) cant be factored... so on this one I will complete the square.

for my zeros:

3x/3 = 0/3 x = 0

AND

(x-2)^2 = 3

X = 2 +/- sqrt(3)


Thanks