Question 363568
Solution : f(X)=2x^3+3x^2-11x-6 
Plug in x = 2 in this
f(2) = 2(2)^3 + 3(2)^2-11(2)-6 = 16+12-22-6=28-28 = 0
so x = 2 is a root
divide the polynomial with (x-2)
Then we get 
2x^2+7x+3
Plug in x = -3 in this
2(-3)^2 +7(-3)+3 
=18-21+3
=0
so x = -3 is another root
and on dividing 2x^2+7x+3 with x+3  we get 2x+1 
and the root of 2x+1 = 0 is  x = -0.5
so the roots are x = 2,-3,-0.5