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

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Answer 1 · 2017-01-02T01:51:58

$- 4, 2 and 3$ $-4, 2 and 3$ .

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

$P (n) = (n - 2) (n^{2} + k n - 12)) .$ $P(n) = (n-2)(n^2+kn-12)).$

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

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

And so, the other two zeros are $- 4 and 3$ $-4 and 3$ .

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What are the possible integral zeros of P(n)=n^3-3n^2-10n+24P(n)=n3−3n2−10n+24P(n)=n^3-3n^2-10n+24?