If it had those zeros, and you solved it, you would end up with
x=-2; x=1; x=2; x=4
and before that you would have had
x+2=0; x-1=0; x-2=0; x-4=0
and before that you would have had
(x+2)(x-1)(x-2)(x-4) = 0
And so the polynomial you would have started with and factored and set
equal to zero, would have been:
P(x) = (x+2)(x-1)(x-2)(x-4)
That's the factored form asked for.
If you multiplied it out it, the polynomial would have been
P(x) = x4 - 5x3 + 20x - 16
real zeros are the same thing as x-coordinates of the x-intercepts:
Edwin
