Question

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

Answer 1

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

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

.