#-2x^3-8x^2+6x+36#
#=-2(x^3+4x^2-3x-18)#
Let #f(x) = x^3+4x^2-3x-18#
By the rational roots theorem, any rational root of #f(x) = 0# of the form #p/q# in lowest terms must be such that #p# is a divisor of #18# and #q# is a divisor of #1#. That is:
#x = +-1#, #+-2#, #+-3#, #+-6#, #+-9# or #+-18#
Trying the first few:
#f(1) = 1+4-3-18 = -16#
#f(-1) = -1+4+3-18 = -12#
#f(2) = 8+16-6-18 = 0#
So #x=2# is a root of #f(x) = 0# and #(x-2)# is a factor of #f(x)#
#x^3+4x^2-3x-18 = (x-2)(x^2+6x+9)#
#x^2+6x+9# is recognisable as a perfect square trinomial:
It is of the form #a^2+2ab+b^2 = (a+b)^2# with #a=x# and #b=3#, so
#(x^2+6x+9) = (x+3)^2#
Putting this all together we get:
#-2x^3-8x^2+6x+36 = -2(x-2)(x+3)(x+3)#
So #-2x^3-8x^2+6x+36=0# has one root #x=2# and one repeated root #x=-3# with multiplicity #2#.
