Question 1173907
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I would be curious to know if the requirements as you show them are really what was intended.  It makes for strange polynomials.<br>
And if -3, -1, and 4 are to be the ONLY zeros, then there are no polynomials that satisfy all the conditions.<br>
The solutions are going to be complicated even in factored form; if you really need them in standard form, I'll let you do that part.<br>
Let's start with leading coefficient 2 and zeros at -3, -1, and 4.  The basic polynomial is<br>
{{{p(x) = -2(x+3)(x+1)(x-4)}}}<br>
{{{graph(400,400,-5,5,-50,50,-2(x+3)(x+1)(x-4))}}}<br>
For that polynomial, f(0)=24; but the requirement is f(0)=5.<br>
To get a polynomial with the given zeros and leading coefficient, and also with f(0)=5, we need to add further linear factors of the form (x+a), (x+b), ..., where the product of ab... is 5/24.<br>
That will make f(0)=(5/24)(24)=5, as required; but of course it will add more zeros to the function.<br>
We can do it with one more factor of (x+5/24):<br>
{{{p(x) = -2(x+3)(x+1)(x-4)(x+5/24)}}}<br>
{{{graph(400,400,-5,5,-50,50,-2(x+3)(x+1)(x-4)(x+5/24))}}}<br>
Or we could do it with two additional linear factors, like (x+5/8) and (x+1/3):<br>
{{{p(x) = -2(x+3)(x+1)(x-4)(x+5/8)(x+1/3)}}}<br>
{{{graph(400,400,-5,5,-50,50,-2(x+3)(x+1)(x-4)(x+5/8)(x+1/3))}}}<br>
I suspect, however, that it was not intended for you to do that kind of thing to answer the question.  I think the statement of the problem simply put too many requirements on the function to make an answer possible.<br>