How do you factor #x^3 + 14x^2 + 60x + 72 = 0#?

1 Answer
Jul 24, 2018

Answer:

#(x+2)(x+6)^2 = 0#

Explanation:

Given:

#x^3+14x^2+60x+72 = 0#

By the rational roots theorem, any rational zeros of the given cubic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #72# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-6, +-8, +-9, +-12, +-18, +-24, +-36, +-72#

In addition, note that all of the coefficients are positive and the constant term is non-zero. As a result, any real zero (rational or otherwise) of this cubic must be negative.

So that leaves rational possibilities:

#-1, -2, -3, -4, -6, -8, -9, -12, -18, -24, -36, -72#

We find:

#(color(blue)(-2))^3+14(color(blue)(-2))^2+60(color(blue)(-2))+72 = -8+56-120+72 = 0#

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

#x^3+14x^2+60+72 = (x+2)(x^2+12x+36)#

Without trying any more of our "possible" zeros, we can recognise the remaining quadratic factor as a perfect square trinomial:

#x^2+12x+36 = x^2+2(x)(6)+6^2 = (x+6)^2#

So the factored form of the given cubic equation can be written:

#(x+2)(x+6)^2 = 0#