Question

Find the zeros of j(x)= 12x^3/5 -27x/5.

Answer 1

To find the zeros of the function \( j(x) = \frac{12x^3}{5} - \frac{27x}{5} \), we set \( j(x) = 0 \) and solve for \( x \):

\[
\frac{12x^3}{5} - \frac{27x}{5} = 0
\]

To eliminate the fraction, we can multiply both sides of the equation by 5:

\[
12x^3 - 27x = 0
\]

Next, we can factor out \( 3x \):

\[
3x(4x^2 - 9) = 0
\]

This equation gives two factors, so we can set each factor to zero:

1. \( 3x = 0 \)
2. \( 4x^2 - 9 = 0 \)

From the first factor, we have:

\[
x = 0
\]

For the second factor, we can solve for \( x \):

\[
4x^2 - 9 = 0
\]

Add 9 to both sides:

\[
4x^2 = 9
\]

Now divide by 4:

\[
x^2 = \frac{9}{4}
\]

Taking the square root of both sides gives us:

\[
x = \pm \frac{3}{2}
\]

So the zeros of the function \( j(x) \) are:

\[
x = 0, \quad x = \frac{3}{2}, \quad x = -\frac{3}{2}
\]

Thus, the final answer is:

\[
\boxed{0, \frac{3}{2}, -\frac{3}{2}}
\]