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:
- \( 3x = 0 \)
- \( 4x^2 - 9 = 0 \)
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:
\[ \boxed{0, \frac{3}{2}, -\frac{3}{2}} \]
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