SOLUTION: Given these key features: f(0) = 5 leading coefficient is -2 zeroes at -3, -1, 4 Create two possible algebraic models, using two different degrees of polynomial functions. Exp

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Question 1173907: Given these key features:
f(0) = 5
leading coefficient is -2
zeroes at -3, -1, 4
Create two possible algebraic models, using two different degrees of polynomial functions.
Express both in standard and factored form.
Answer by greenestamps(13215) About Me  (Show Source):
You can put this solution on YOUR website!


I would be curious to know if the requirements as you show them are really what was intended. It makes for strange polynomials.

And if -3, -1, and 4 are to be the ONLY zeros, then there are no polynomials that satisfy all the conditions.

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.

Let's start with leading coefficient 2 and zeros at -3, -1, and 4. The basic polynomial is

p%28x%29+=+-2%28x%2B3%29%28x%2B1%29%28x-4%29

graph%28400%2C400%2C-5%2C5%2C-50%2C50%2C-2%28x%2B3%29%28x%2B1%29%28x-4%29%29

For that polynomial, f(0)=24; but the requirement is f(0)=5.

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.

That will make f(0)=(5/24)(24)=5, as required; but of course it will add more zeros to the function.

We can do it with one more factor of (x+5/24):

p%28x%29+=+-2%28x%2B3%29%28x%2B1%29%28x-4%29%28x%2B5%2F24%29

graph%28400%2C400%2C-5%2C5%2C-50%2C50%2C-2%28x%2B3%29%28x%2B1%29%28x-4%29%28x%2B5%2F24%29%29

Or we could do it with two additional linear factors, like (x+5/8) and (x+1/3):

p%28x%29+=+-2%28x%2B3%29%28x%2B1%29%28x-4%29%28x%2B5%2F8%29%28x%2B1%2F3%29



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.