[tex]f(x)=2x^3-5x^2+x+2[/tex][tex]f(2)=2(2)^3-5(2)^2+(2)+2=16-20+2+2=0[/tex]
Hence, 2 is a zero of f(x). That is x - 2 is a factor of f(x).
So we can find
[tex]\frac{2x^3-5x^2+x+2}{x-2}[/tex][tex]\Rightarrow\frac{2x^3-5x^2+x+2}{x-2}=2x^2-x-1[/tex][tex]\text{Next we solve }2x^2-x-1=0[/tex][tex]\begin{gathered} \Rightarrow2x^2-x-1=0 \\ 2x^2+x-2x-1=0 \\ x(2x+1)-1(2x+1)=0 \\ \Rightarrow(x-1)(2x+1)=0 \\ \Rightarrow x-1=0\text{ or 2x+1=0} \\ \Rightarrow x=1\text{ or 2x=-1} \\ x=1\text{ or x =-}\frac{1}{2} \end{gathered}[/tex]
Hence,
[tex]x=2,1,\text{ or -}\frac{1}{2}[/tex]
