Question

One zero of the polynomial $3x^3+16x^2+15x-18$ is $\frac{2}{3}$. Find the other zeroes of the polynomial.

Answer 1

Given: $x=\frac{2}{3}$ is one of the zero of $3x^3+16x^2+15x-18$
Now, we have
$$\begin{gathered} x=\frac{2}{3} \\ \Rightarrow x-\frac{2}{3}=0 \end{gathered}$$
Now, we divide $3x^3+16x^2+15x-18$ by $x-\frac{2}{3}$ to find the quotient



So, the quotient is $3x^2+18x+27$
Now,
$$\begin{gathered} 3x^2+18x+27=0 \\ \Rightarrow 3x^2+9x+9x+27=0 \\ \Rightarrow 3x\left(x+3\right)+9\left(x+3\right)=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow \left(x+3\right)\left(3x+9\right)=0 \\ \Rightarrow \left(x+3\right)=0\mathrm{or}\left(3x+9\right)=0 \\ \Rightarrow x=-3\mathrm{or}x=-3 \end{gathered}$$