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

##### 1 Answer

#### Explanation:

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

By the rational roots theorem, any *rational* zeros of

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:

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

To factor the remaining quadratic note that

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)#