#f(x) = x^3 +2x^2 -23x-60#
There are 4 terms in the expression to be factorised, we could consider pairing them, but #23# is a prime number and therefore has no factors other than #1 and 23#, and 23 is not a factor of 60.
Let's use the factor theorem:
The factors of #60# are : # +-1, +-2, +-3, +-4, +-5, +-6, +-10, +-12, +-15, +-20, +-30,+-60#
But we can try the prime factors first:
#f(2) = 2^3 +2(2)^2 -23(2)-60 = -90#
#f(-2) = (-2)^3 +2(-2)^2 -23(-2)-60 = -14#
#f(3) = 3^3 +2(3)^2 -23(3)-60 = -84#
#f(-3) = (-3)^3 +2(-3)^2 -23(-3)-60 = color(red)(0)#
#rarr :.color(red)( (x+3))# is a factor
#f(5) = 5^3 +2(5)^2 -23(5)-60 = color(red)(0)#
#rarr :.color(red) ((x-5))# is a factor
If #-3 and 5# give #0#, let's try #-4# because #3 xx-5 xx4= -60#
#f(-4) = (-4)^3 +2(-4)^2 -23(-4)-60 = color(red)(0)#
#rarr :. color(red)((x+4))# is a factor.
Once we had found the first factor, we could also have used long division or synthetic division to find a trinomial and then factorised that.
#f(x)= (x+3)(x-5)(x+4)#
Check:
#(x+3)(x^2-x-20) = x^3-x^2-20x+3x^2-3x-60#
#=x^3+2x^2-23x -60# so, it all checks out fine!
