Solution:
Given the expression below
[tex]f(x)=2x^3-5x^2-9x+18[/tex]
And x + 2 is a factor
Applying long division
[tex](2x^3-5x^2-9x+18)\div(x+2)[/tex]
After the long divison, factorizing the quotient gives,
[tex]\begin{gathered} 2x^2-9x+9 \\ =2x^2-6x-3x+9 \\ =2x(x-3)-3(x-3) \\ =(2x-3)(x-3) \end{gathered}[/tex]
The factored form of the expression is
[tex]f(x)=(x+2)(2x-3)(x-3)_{}[/tex]
To find the zeros, equating each factor to zero, i.e
[tex]\begin{gathered} (x+2)(2x-3)(x-3)_{}=0 \\ x+2=0 \\ x=-2 \\ 2x-3=0 \\ 2x=3 \\ \frac{2x}{2}=\frac{3}{2} \\ x=\frac{3}{2} \\ x-3=0 \\ x=3 \\ x=-2,\frac{3}{2},3 \end{gathered}[/tex]
Hence, the zeros of f(x) are
[tex]x=-2,\frac{3}{2},3[/tex]
