Let's start by looking for some quantity that appears on every term: since every term has the variable #m# in it, we can factor it out. More specifically, since the powers of #m# are the fifth, the fourth and the third, every term contains #m^3#. If we factor it out, we get
#-63m^5+27m^4-18m^3 = m^3(-63m^2+27m-18)#
Now we can look at the numerical coefficients. Since #63=3^2*7#, #27^3^3# and #18=2*3^2#, we can observe that every coefficient contains a #3^2=9#, and so we can factor it as well:
#m^3(-63m^2+27m-18)=9m^3(-7m^2+3m-2)#
The only possible way to further factor this expression is to solve the quadratic expression in the parenthesis, but unfortunately its determinant #\Delta = b^2-4ac# is negative, so the expression has no solutions, and can't be factored in the form #(x-x_1)(x-x_2)#, where #x_1# and #x_2# would have been the solutions.
