What are the zeros of the polynominal?
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If we can reduce this to a quadratic by finding an easy factor, then we can find the zeroes. If we substitute a set of values of x and work out the result for p(x), we may be able to work out a factor. Let's start with x=0, then p(0)=60. Let's try x=1. p(1)=36 which is smaller than 60, so we're on track for zero. p(2)=14, smaller again. p(3)=27-18-69+60=0. Bingo! So (x-3) is a factor. Divide the cubic by (x-3) using synthetic division and we get x^2+x-20 which, surprise, surprise, factorises into (x+5)(x-4). We have th three zeroes: x=-5, 3 and 4. Let's check out -5: -125-50+115+60=0. Success!

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