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: ( − 2 ) 3 + 14 ( − 2 ) 2 + 60 ( − 2 ) + 72 = − 8 + 56 − 120 + 72 = 0 So x = − 2 is a zero and ( x + 2 ) a factor: x 3 + 14 x 2 + 60 + 72 = ( x + 2 ) ( x 2 + 12 x + 36 ) Without trying any more of our "possible" zeros, we can recognise the remaining quadratic factor as a perfect square trinomial: x 2 + 12 x + 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
