Given:
[tex]f(x)=3x^3-8x^2-5x+6[/tex]
c= -1 is the zero of the function.
Use the synthetic division,
Solving it further,
[tex]\begin{gathered} (x-1)(3x^2-11x+6)=0 \\ \text{Solve }3x^2-11x+6=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a},a=3,b=-11,c=6 \\ x=\frac{-\left(-11\right)\pm\sqrt{\left(-11\right)^2-4\cdot\:3\cdot\:6}}{2\cdot\:3} \\ x=\frac{11\pm7}{6} \\ x=\frac{11+7}{6},x=\frac{11-7}{6} \\ x=3,x=\frac{2}{3} \end{gathered}[/tex]
So, the zeros of the function are,
[tex]\begin{gathered} f(x)=3x^3-8x^2-5x+6 \\ x=-1,3,\frac{2}{3} \\ f(x)=(x+1)(3x-2)(x-3) \end{gathered}[/tex]
Answer:
