Given:
[tex]3x^3+x^2-20x+12[/tex]
Factor: x+3
To find the zeros:
Using synthetic division,
So, the polynomial can be written as,
[tex]3x^3+x^2-20x+12=(x+3)(3x^2-8x+4)[/tex]
Let us consider,
[tex]p(x)=3x^2-8x+4[/tex]
Put x=2, we get
[tex]\begin{gathered} p(2)=3(2^2)-8(2)+4 \\ =12-16+4 \\ =0 \end{gathered}[/tex]
Hence, x=2 is the other one zero of the polynomial.
Put x=2/3, w get
[tex]\begin{gathered} p(\frac{2}{3})=3(\frac{2}{3})^2-8(\frac{2}{3})+4 \\ =\frac{4}{3}-\frac{16}{3}+4 \\ =-\frac{12}{3}+4 \\ =\frac{-12+12}{3} \\ =0 \end{gathered}[/tex]
Hence, x=2/3 is the other one zero of the polynomial.
Hence, the zeros of the polynomial are,
[tex]-3,\frac{2}{3},\text{ and, 2}[/tex]
