Question

find the zeroes of j(x)=12x^3/5 - 27x/5

Answer 1

To find the zeroes of the function \( j(x) = \frac{12}{5}x^3 - \frac{27}{5}x \), we need to set the function equal to zero and solve for \( x \):

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

We can factor out \( \frac{1}{5} \) from the equation:

\[
\frac{1}{5}(12x^3 - 27x) = 0.
\]

Since \( \frac{1}{5} \) does not affect the zeroes, we can simplify to:

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

Next, factor out \( 3x \):

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

Now, set each factor equal to zero:

1. \( 3x = 0 \) gives \( x = 0 \).
2. \( 4x^2 - 9 = 0 \) leads to:

\[
4x^2 = 9 \implies x^2 = \frac{9}{4} \implies x = \pm \frac{3}{2}.
\]

Thus, the zeroes of the function \( j(x) \) are:

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

In summary, the zeroes of \( j(x) \) are \( x = 0, \frac{3}{2}, -\frac{3}{2} \).