Question

What are the zeros of j(x) in the polynomial equation?

Answer 1

Answer: x =
`(-3)/(2)`, 0,
`(3)/(2)`

Explanation:

To find the zeros, we will find the solutions to this equation that make it equal 0. This is a cubic function, so we can assume that there will be three solutions.

Given:

`\displaystyle J(x) = (12x^3)/(5) -(27x)/(5)`

Set the function equal to 0:

`\displaystyle 0 = (12x^3)/(5) -(27x)/(5)`

Subtract:

`\displaystyle 0 = (12x^3-27x)/(5)`

Multiply both sides of the equation by 5:

`\displaystyle 0 = 12x^3-27x`

Factor the polynomial:

`\displaystyle 0 = (3x)(2x-3)(2x+3)`

Solve with the zero product property:

0 = 3x 0 = 2x - 3 0 = 2x + 3

0 = x 3 = 2x -3 = 2x

x = 0 x =
`(3)/(2)` x =
`(-3)/(2)`