Find a possible formula for the polynomial \(g(x)\) that is 4th degree, has a double zero at \(x=3\), \(g(5)=0 \), \(g(-1)=0\), and \(g(0)=3\)

5th Ed: #21

Since \(g(x)\) has a double zero at \( x=3\), that means we have a factor of \( (x-3)^2\).

Since \(g(5)=0 \) and \(g(-1)=0\), we have factors \( (x-5)(x+1) \)

This gives us \( g(x)=a(x-3)^2(x-5)(x+1) \)

Since \(g(0)=3 \) we have, \( 3=a(-3)^2(-5)(1) \), so \( a=\frac{-1}{15} \)

Therefore, \( g(x)=\frac{-1}{15}(x-3)^2(x-5)(x+1) \)