I need to find an expression that satisfies the qualifying conditions for a quintic polynomial.
$f(0)=3$ and $f(-2)=f(\frac{1}{2})=f(1)=0$.
With this information, I found that the zeros are $2, -\frac{1}{2},$ and $-1$.
By plugging $0$ into $f(x)$, I found that $F=3$ using the form $ax^5+bx^4+cx^3+dx^2+ex+f$.
Any advice where to go from here?
$\endgroup$2 Answers
$\begingroup$One set of possibilities is ... $$ f(x) = k \left [ (x+2)^n(x-\frac12)^p(x-1)^q \right ]$$ where $n,p,q$ are positive integers that sum to $5$
Choose any three you like and then use $f(0)=3$ to calculate $k$
$\endgroup$ 1 $\begingroup$Plugging in $x=-2$, $1/2$ and $1$ into $f(x)$ will give you three linear equations in the four remaining unknown parameters.
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