Constructing a polynomial with certain zeroes. I want to construct a polynomial $f(x)$ that has zeroes at $-9,\,-5,\,0,\, 5,\, 9$.
Can somebody provide a method (or perhaps some hints) for solving this?
 A: And nowhere else, I guess, otherwise the zero polynomial will do. And I assume your are working over the rationals, say.
Just use the fact that $f(x)$ has a zero at $a$ if and only if $x - a$ divides $f(x)$.
A: Zeroes at $-9$, $-5$, $0$, $5$, $9$... This means that the polynomial will be of the form:
$$f(x)=a(x)(x+9)(x+5)(x-5)(x-9)$$
Where $a$ is a constant. Any value of $a$ will work here. These are all correct answers:
$$f(x)=x(x+9)(x+5)(x-5)(x-9)$$
$$f(x)=\sqrt 2(x)(x+9)(x+5)(x-5)(x-9)$$
$$f(x)=\pi x(x+9)(x+5)(x-5)(x-9)$$
How did I construct this? The zeroes are $-9$, $-5$, $0$, $5$, $9$, which means that these $5$ values of $x$ will make $f(x)$ equal to $0$. Anything times $0$ will make the product $0$. These $5$ values will make $1$ factor equal to $0$, which makes $f(x)=0$.
A: If you want a polynomial that lies in $\mathbb{Z}[x]$ consider $f(x)=x(x+9)(x-9)(x-5)(x+5)$. This is the polynomial with the minimal degree such that it has zeros at $0,\pm 9,\pm 5$.
A: Try this:
$$f(x):=(x-5)(x+5)(x-9)(x+9)x$$
