Answer
$f(x)=-\displaystyle \frac{1}{2}x^{3}+2x^{2}+\frac{11}{2}x-15$
Work Step by Step
For any polynomial function $f(x),\ (x-k)$ is a factor of the polynomial if and only if $f(k)=0$.
So, we can write
$ f(x)=a(x-2)(x+3)(x-5)\quad$ for some number a.
To find $a,$ use the given information: $f(3)=6$
$f(2)=a(3-2)(3+3)(3-5)=6$
$a(1)(6)(-2)=6$
$-12a=6$
$a=-1/2$
So,
$f(x)=-\displaystyle \frac{1}{2}(x-2)(x+3)(x-5)$
Rewrite in standard form
$f(x)=-\displaystyle \frac{1}{2}(x-2)(x+3)(x-5)$
$f(x)=-\displaystyle \frac{1}{2}(x-2)(x^{2}-2x-15)$
$f(x)=-\displaystyle \frac{1}{2}[x(x^{2}-2x-15) -2(x^{2}-2x-15)]$
$f(x)=-\displaystyle \frac{1}{2}[x^{3}-2x^{2}-15x-2x^{2}+4x+30]$
$f(x)=-\displaystyle \frac{1}{2}[x^{3}-4x^{2}-11x+30]$
$f(x)=-\displaystyle \frac{1}{2}x^{3}+2x^{2}+\frac{11}{2}x-15$
