SOLUTION: Hello. Can you help me figure out what I am doing wrong? I need to find the Zeros of the following function. {{{ f(x)=3x^5-9x^3 }}} I am getting 0(multiplicity 3), -3,

Question 233189: Hello.

Can you help me figure out what I am doing wrong?
I need to find the Zeros of the following function.

I am getting 0(multiplicity 3), -3, 3.
Thank you in advance.
William
Found 2 solutions by edjones, Theo:
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
3x^5-9x^3=0
3x^3(x^2-3)=0
3x^3=0
x=0
.
x^2=3
x=+-sqrt(3)
.
The zeros are 0, sqrt(3), -sqrt(3)
.
Ed
.

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
y = 3x^5 - 9x^3

Set the equation equal to 0 to get:

3x^5 - 9x^3 = 0

factor out x^3 to get:

x^3 * (3x^2 - 9) = 0

This means that x^3 = 0 or (3x^2-9) = 0 or both.

If x^3 = 0, then x = root(3,0) = 0

If (3x^2-9) = 0, then:

Divide by 3 to get:

x^2 - 3 = 0

Add 9 to both sides to get:

x^2 = 3

Take square root of both sides to get:

x = +/- sqrt(3)

Your answers should be that x = 0 or x = sqrt(3) or x = -sqrt(3)

Plug these values into the original equation to see if the equation is true.

With x = 0, 3x^5-9x^3 becomes 0 so this part is true.

With x = sqrt(3), 3x^5-9x^3 becomes 3*(sqrt(3))^5 - 9*(sqrt(3))^3

This becomes 3*3*3*sqrt(3) - 9*3*sqrt(3) = 27*sqrt(3) - 27*sqrt(3) = 0 so this part is true.

With x = -sqrt(3), 3x^5 - 9x^3 becomes 3 * (-sqrt(3))^5 - 3*(-sqrt(3))^3

This becomes 3*3*3*(-sqrt(3)) - 9*3*(-sqrt(3)) = -27*(sqrt(3)) + (27*(sqrt(3) = 0 so this part is true.

Your answers are:

0,-sqrt(3),sqrt(3)

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