Question

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

Answer 1

`(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`