# What are the possible integral zeros of P(n)=n^3-3n^2-10n+24?

Jan 2, 2017

$- 4 , 2 \mathmr{and} 3$.

#### Explanation:

P(2)=0. So, $n - 2$ is a factor. Now,

P(n) = (n-2)(n^2+kn-12)).

Comparing coefficient of ${n}^{2} = k - 2$ with $- 3$, k = -1.

So, $P \left(n\right) = \left(n - 2\right) \left({n}^{2} - n - 12\right) = \left(4 - 2\right) \left(n + 4\right) \left(n - 3\right)$.

And so, the other two zeros are $- 4 \mathmr{and} 3$.

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