Question 1177496
<pre>
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.

{{{drawing(400,600,-4,5,-21,14,
graph(400,600,-4,5,-21,14,(x+3)(x-4)(x-1/12)),locate(0,1.5,"(0,1)") )}}}

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

{{{p(x) = (x+3)(x-4)(x-a)}}}

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

{{{p(0) = (0+3)(0-4)(-a)=1}}}

{{{(3)(-4)(-a)=1}}}

{{{12a=1}}}

{{{a=1/12}}} <--that is between -3 and 4

So such a polynomial would be

{{{p(x) = (x+3)(x-4)(x-1/12)}}}
 
If you multiply that out, you get:

{{{p(x) = x^3 - expr(13/12)x^2 - expr(143/12)x + 1}}}

That's what the graph above is of.

Edwin</pre>