By the rational root theorem, all rational roots of #P(x)# must be of the form #p/q# in lowest terms for some integers #p#, #q# where #p# is a divisor of the constant term #36# of #P(x)# and #q# is a divisor of the coefficient #1# of the highest order term #x^4#.
#36 = 2^2*3^2#
So any rational roots must be one of:
#+-1#, #+-2#, #+-3#, #+-4#, #+-6#, #+-9#, #+-12#, #+-18# or #+-36#
#P(1) = 1-1-15+3+36 = 24#
#P(-1) = 1+1-15-3+36 = 20#
#P(2) = 16-8-60+6+36 = -10#
#P(-2) = 16+8-60-6+36 = -6#
#P(3) = 81-27-135+9+36 = -36#
#P(-3) = 81+27-135-9+36 = 0#
#P(4) = 256-64-240+12+36 = 0#
So #x = -3# and #x = 4# are roots and #(x+3)# and #(x-4)# are factors.
If we divide #P(x)# by #(x+3)(x-4)# then the constant term of the resulting polynomial will be #-3#. So the only possible remaining rational roots would be #+-1# or #+-3#. We have eliminated #+-1# and #3#, but it is possible that #-3# is a repeated root? No, because if it were, it would be possible to divide by #(x+3)# again, leaving a linear factor #(x-1)# and we already know that #1# is not a root.
So we have found all of the rational roots. What else is there?
Let's divide #P(x)# by #(x+3)(x-4)# to find out ...
#(x+3)(x-4) = x^2-x-12#
#P(x) = x^4-x^3-15x^2+3x+36#
#=(x^4-x^3-12x^2)-(3x^2-3x-36)#
#=x^2(x^2-x-12)-3(x^2-x-12)#
#=(x^2-3)(x^2-x-12)#
So #(P(x))/(x^2-x-12) = x^2-3#
which has roots #x = +-sqrt(3)#
