By the rational root theorem, any rational roots of #P(x) = 0# must be expressible in the form #p/q#, where #p, q in ZZ#, #q > 0#, #"hcf"(p, q) = 1#, #p# a divisor of the constant term #-100# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational roots of #P(x) = 0# are:

#+-1#, #+-2#, #+-4#, #+-5#, #+-10#, #+-25#, #+-50#, #+-100#

Let's try some:

#P(1) = 1+6-15-100 = -108#

#P(-1) = -1+6+15-100 = -80#

#P(2) = 8+24-30-100 = -102#

#P(-2) = -8+24+30-100 = -54#

#P(4) = 64+96-60-100 = 0#

So #(x-4)# is a factor of #P(x)#.

Use synthetic division to divide #P(x)# by #(x-4)#...

So #x^3+6x^2-15x-100 = (x-4)(x^2+10x+25)#

Now we can recognise #x^2+10x+25# as a perfect square trinomial. It is of the form #a^2+2ab+b^2 = (a+b)^2#, with #a = x# and #b = 5#.

So #x^2+10x+25 = (x+5)^2#

Putting this together:

#x^3+6x^2-15x-100 = (x-4)(x+5)(x+5)#