Question

Can `y=x^3+5x^2-12x-36` be factored? If so what are the factors ?

Answer 1

`x^3+5x^2-12x-36 = (x-3)(x+2)(x+6)`

Explanation:

`f(x) = x^3+5x^2-12x-36`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-36` 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, +-9, +-12, +-18, +-36`

In addition note that the signs of the coefficients are in the pattern: `+ + - -`. By Descartes' rule of signs we can deduce that `f(x)` has exactly `1` positive Real zero. It has `0` or `2` negative Real zeros.

So let's try the positive possibilities first:

`f(1) = 1+5-12-36 = -42`

`f(2) = 8+20-24-36 = -32`

`f(3) = 27+45-36-36 = 0`

So `x=3` is a zero and `(x-3)` a factor:

`x^3+5x^2-12x-36 = (x-3)(x^2+8x+12)`

To factor the remaining quadratic note that `2+6=8` and `2*6=12`

Hence:

`x^2+8x+12 = (x+2)(x+6)`

So our complete factorisation is:

`x^3+5x^2-12x-36 = (x-3)(x+2)(x+6)`