Question 1147361
<pre>
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) = x<sup>4</sup> - 5x<sup>3</sup> + 20x - 16

real zeros are the same thing as x-coordinates of the x-intercepts:

{{{graph(3200/19,800,-3,5,-35,3,x^4 - 5x^3 + 20x - 16)}}} 

Edwin</pre>