# Finding polynomial function with given zeros

1. Oct 6, 2012

### NneO

1. The problem statement, all variables and given/known data
find the polynomial fucntion p(x) with zeros, -1, 1, 3 and P(0)=9

2. Relevant equations
all i have is (x^2-1) and (x-3)

3. The attempt at a solution

2. Oct 6, 2012

### Mentallic

If a polynomial has roots a,b,c for example, then the simplest polynomial that satisfies this problem is $p(x)=(x-a)(x-b)(x-c)$

So the simplest polynomial that satisfies the roots -1,1,3 is $p(x)=(x+1)(x-1)(x-3)$ which is a cubic. But if we evaluate p(0) we get $p(0)=(0+1)(0-1)(0-3)=3$ but we were given that p(0)=9, so what can we do?

3. Oct 6, 2012

### HallsofIvy

Staff Emeritus
Does the problem really say "find the polynomial function" (emphasis mine)? The clearly is NOT a single such a polynomial. There does exist a "simplest" polynomial, or "polynomial of least degree" satisfying those conditions but give such a polynomial we could always multiply by other polynomials to get a higher order polynomial satisfying those conditions.

4. Oct 6, 2012

### SammyS

Staff Emeritus
Hello NneO. Welcome to PF !

As you can deduce from Mentallic and HallsofIvy, there are many polynomials which can satisfy the stated requirements. Any polynomial which does satisfy those requirements, will have the factors (x-1), (x+1), and (x-3) as you indicated.

There are a few ways to get the desired value for P(0). The easiest is as Mentallic suggested.

You can also introduce an additional (real) zero which could do the trick. That might not be allowed, depending upon how the problem is interpreted. You could introduce a factor, such as (x2 + a), which has no real zeros, if a>0 .