According to the context, the plot of the polynomial should be located in the 2nd and 4th quadrants 

with the y-intercept at y= 1,


with no parts in the 1st and/or 3rd quadrant.



A polynomial satisfying such conditions DOES NOT EXIST.

It doesn't say "with no parts in the 1st and/or 3rd quadrant", it just
says "extending from quadrant 2 to quadrant 4".  It can extend from quadrant
2 to quadrant 4 by passing through quadrant 1 on its way to quadrant 4. 

It might very well look like the graph  below, with two turning points,
and having zeros at, arbitrarily, -3 and 4, and going up on the extreme
right.  So we'll make sure it has a positive leading term.



To have 2 turning points, it must have degree 3 or more.  So let's try to
make it have 3 real zeros.

To have a zero at -3 it must have factor (x+3)

To have a zero at 4 it must have factor (x-4)

It must have another zero between -3 and 4. Let's suppose that third zero
has value a.  So the polynomial must also have a factor (x-a)

It would have degree 3 if it were of the form



and hopefully "a" will turn out to be between -3 and 4.

In order for the polynomial to have y-intercept 1, it must go through (0,1).

So we substitute x=0 and p(0)=1







 <--that is between -3 and 4

So such a polynomial would be


 
If you multiply that out, you get:



That's what the graph above is of.

Edwin