One zero of the polynomial

Solution:

Let $f(x)=3 x^{3}+16 x^{2}+15 x-18$

It is given that one of its zeroes is $\frac{2}{3}$.

Therefore, one factor of $f(x)$ is $\left(x-\frac{2}{3}\right)$

We get another factor of $f(x)$ by dividing it with $\left(x-\frac{2}{3}\right)$.

On division, we get the quotient $3 x^{2}+18 x+27$

$\Rightarrow f(x)=\left(x-\frac{2}{3}\right)\left(3 x^{2}+18 x+27\right)$

$=\left(x-\frac{2}{3}\right)\left(3 x^{2}+9 x+9 x+27\right)$

$=\left(x-\frac{2}{3}\right)(3 x(x+3)+9(x+3))$

$=\left(x-\frac{2}{3}\right)(3 x+9)(x+3)$

$=3\left(x-\frac{2}{3}\right)(x+3)(x+3)$

$=3\left(x-\frac{2}{3}\right)(x+3)^{2}$

To find the zeroes, we put $f(x)=0$

$\Rightarrow 3\left(x-\frac{2}{3}\right)(x+3)^{2}=0$

$\Rightarrow\left(x-\frac{2}{3}\right)=0$ or $(x+3)^{2}=0$

$\Rightarrow x=\frac{2}{3},-3$

Hence, other zero of the polynomial f(x) is –3.