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

Question

2008 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-02T01:51:58

$-4, 2 and 3$ .

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

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

.

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