We have to find the roots of P(x) = x^3 - 3x^2 - 10x + 24.

Also we are given that x = 2 is a root of P(x)

Now x^3 - 3x^2 - 10x + 24 = x^3 - 2x^2 - x^2 + 2x - 12x + 24

=> x^2(x - 2) - x( x- 2) - 12(x - 2) = 0

=> (x - 2)(x^2 - x - 12) = 0

=> (x - 2)( x^2 - 4x + 3x - 12) = 0

=> (x - 2) ( x(x - 4) + 3(x - 4)) = 0

=> (x - 2)( x + 3) (x - 4) = 0

**Therefore the roots of P(x) = x^3 - 3x^2 - 10x + 24 other than x = 2 are x = -3 and x = 4.**

Given the polynomial P(x) = x^3 - 3x^2 - 10x + 24

We are given that x=2 is one of the root.

Then we know that (x-2) is one of the factors of P(x).

==> P(x) = (x-2) * Q(x)

Now we will divide P(x) / (x-2) to determine Q(x).

==> P(x) = (x-2) *(x^2 -x -12)

Now we will factor the binomial.

==> P(x) = (x-2)(x-4)(x+3)

==> x1 = 2 x2= 4 x3= -3

**Then, the zeros of the polynomial P(x) are:**

**x = { -3, 2, 4}**