Answer
-2, -1, 2, and 3 of multiplicity one
See graph below.
Work Step by Step
The question asks for all real zeros of P(x), their respective multiplicities, and the graph of P(x)
Given $P(x) = x^4 - 2x^3 - 7x^2 + 8x + 12$
$P(x) = x^4 - 4x^2 - 2x^3 +8x - 3x^2 + 12$
$P(x) = x^2 (x^2 - 4) - 2x (x^2 - 4) - 3(x^2 - 4)$
$P(x) = (x^2 - 2x - 3) (x^2-4)$
$P(x) = (x-3)(x+1)(x-2)(x+2)$
Set $P(x) = 0$
Thus x = -2, -1, 2, 3
To determine multiplicity, it is the power of the root determined from setting P(x) = 0
So -2, -1, 2, and 3 of multiplicity one
See graph below.
