one factor of p(x) = 2x^3-9x^2-11x+30 please solve it

To solve: p(x) =
2x^3-9x^2-11x+30

Solution.

P(-2)
= 2(-2)^3-9(-2)^2-11(-2)+30

= -16-36
-11(-2)+30

= -52+52=0

So ,
x-(-2) = x+2 is a factor.

Therefore p(x )= p(x) =
2x^3-9x^2-11x+30= (x+2)(ax^2+bx+c). But a and c could easily guessed by equating  x^3
and constant terms: a= 2 and c = 15

Therefore,  p(x) =
2x^3-9x^2-11x+30 = (x+2)(2x^2+bx+15) = 0

Equating the
coefficients of x on both sides,

-11 = 15x+2k
Or

k = (-11-15)/2 = -13.

So
the required factors of p(x) are: (x+2) and
2x^2-13x+15.

But 2x^2-13x+15 = 2x^2
-10x-3x+15

=2x(x-5)-3(x-5)

=(x-5)(2x-3),
zeros are x=5 and x=3/2.

Therefore the zeros of p(x) are 
-2,  3/2 and 5, or are the solutions of p(x) = 0.