P= x^3 + mx^2 + 3mx -9

P is dividible by (x-3)

Then (x-3) is one of P's factors, and 3 is one of the roots for P.

==> P(x) = (x-3)*Q(x)

==> P(3) = 0

P(3) = 3^3 + m(3^2) + 3m(3) - 9 = 0

==> 27 + 9m + 9m - 9 = 0

==> 18 + 18m = 0

==> 18m = -18

==> m = -1

P(x) = x^3+mx^2+3mx-9

If P(x) is divisible by x-3, then by remainder theorem P(3) = 0

So P(3) = 3^3-m*3^2+3m*3-9 = 0

27 - 9m^2+9m -9 = 0. Divide by 9

3-m^2+m-1 = 0

-m^2+m+2 = 0. Multiply by -1.

m^2-m-2 = 0

m^2-2m+m-2 = 0

m(m-2)+1(m-2) = 0

(m-2)(m+1) = 0

m-2 = 0 or m+1 =0

So m=2 or m = -1.

If the polynomial is divisible by (x-3), then x=3 verify the equation:

P(3)=0

This equation is the result of applying the rule of the division with reminder:

P(x) = (x-3)*Q(x)

The reminder is R=0, because P is divisible by x-3.

Now, we'll substitute x by 3 and we'll get:

P(3) = (3-3)*Q(3)

P(3) = 0

P(3) = 3^3 + m*3^2 + 3*m*3 - 9

P(3) = 27 + 9m + 9m - 9

We'll combine like terms:

18m + 18 = 0

We'll factorize:

18 (m+1) = 0

We'll divide by 18:

m+1 = 0

We'll subtract 1 both sides:

**m=-1**