Step 1
Given;
[tex]g(x)=3x^2-5x-2[/tex]
Required; To find the zeroes by factoring
Step 2
Find two factors that when added gives -5x and when multiplied give -6x
[tex]\begin{gathered} \text{These factors are;} \\ -6x\text{ and x} \end{gathered}[/tex][tex]\begin{gathered} -6x\times x=-6x^2 \\ -6x+x=-5x \end{gathered}[/tex]
Factoring we have and replacing -5x with -6x and x we have
[tex]\begin{gathered} 3x^2-6x+x-2=0 \\ (3x^2-6x)+(x-2)_{}=0 \\ 3x(x-2)+1(x-2)=0 \\ (3x+1)(x-2)=0 \\ 3x+1=0\text{ or x-2=0} \\ x=-\frac{1}{3},2 \\ \text{The z}eroes\text{ are, x=-}\frac{1}{3},2 \end{gathered}[/tex]
Graphically the x-intercepts are;
The x-intercepts are;-1/3,2
Hence, the answer is the zeroes and x-intercepts are the same, they are;
[tex]-\frac{1}{3},2[/tex]
