# What are the possible integral zeros of P(s)=s^3+4s^2-15s-18?

Nov 24, 2017

$\pm 1 , \pm 2 , \pm 3 , \pm 6 \mathmr{and} \pm 9$

#### Explanation:

The possible integral zeros of the equation will be the divisors of the isolated number divided by the coefficient of the largest degree in the polynomial:

The divisors of 18 are $\pm 1 , \pm 2 , \pm 3 , \pm 6 \mathmr{and} \pm 9$

Of course, you should divide these numbers by 1, which is the coefficient of the highest degree member of the polynomial.

The question is not answered until we find the values of the divisors that when substituted in the function it equates to $P \left(s\right) = 0$.

So;
$P \left(- 1\right) = {\left(- 1\right)}^{3} + 4 {\left(- 1\right)}^{2} - 15 \left(- 1\right) - 18 = 0$
$P \left(+ 1\right) = {\left(1\right)}^{3} + 4 {\left(1\right)}^{2} - 15 \left(1\right) - 18 = - 28$ .... and so on.
The result should be $P \left(s\right) = 0$
So the numbers that result in $P \left(s\right) = 0$ from the divisors of 18 are $3 , - 1 , - 6$
That is; P(-1) = 0; P(3) = 0; P(-6) = 0