\(p(x) = 6x^3 + 31x^2 + 4x-5\\ p(-5)=0 \Rightarrow (x+5) \text{ is a factor}\)
\(\dfrac{p(x)}{x+5} = 6 x^2+x-1\\ \text{we can factor this by eye}\\ 6 x^2+x-1 = (3x-1 )(2x +1)\\ p(x) = (3x-1)(2x+1)(x+5)\\ \text{with zeros at }x = \dfrac 1 3,~-\dfrac 1 2,~-5\)
Errr.. what happened... Did you completely change the question?
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