We will have the following:
12. The zeros we can see are located at:
[tex]\begin{gathered} (2-x)(3-2x)=0 \\ \\ \Rightarrow2-x=0\Rightarrow x=2 \\ \\ and \\ \\ \Rightarrow3-2x=0\Rightarrow2x=3\Rightarrow x=\frac{3}{2} \end{gathered}[/tex]
[tex]\begin{gathered} x=2 \\ \\ x=\frac{3}{2} \end{gathered}[/tex]
13. We determine the zeros as follows:
[tex]\begin{gathered} f(x)=9x^2+6x+1\Rightarrow f^{\prime}(x)=18x+6=0 \\ \\ \Rightarrow18x=-6\Rightarrow x=-\frac{1}{3} \end{gathered}[/tex]
So, there is only one zero at:
[tex]x=-\frac{1}{3}[/tex]
