How do you factor #3x^3 - 9x^2 - 54x + 120#?

1 Answer
Aug 18, 2016

Answer:

#3x^3-9x^2-54x+120 = 3(x+2)(x-5)(x+4)#

Explanation:

#3x^3-9x^2-54x+120 = 3(x^3-3x^2-18x+40)#

By the rational roots theorem, any rational zeros of #x^3-3x^2-18x+40# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #40# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-4, +-5, +-8, +-10, +-20, +-40#

Trying each in turn, with #x=2# we get:

#x^3-3x^2-18x+40=8-3(4)-18(2)+40 = 8-12-36+40 = 0#

So #x=2# is a zero and #(x-2)# a factor:

#x^3-3x^2-18x+40 = (x-2)(x^2-x-20)#

To factor the remaining quadratic, note that #5-4=1# and #5*4=20#.

Hence:

#x^2-x-20 = (x-5)(x+4)#

Putting it all together:

#3x^3-9x^2-54x+120 = 3(x+2)(x-5)(x+4)#