Suppose p is a polynomial of degree 3 with three zeros at x=1, x=2, and x=3. Find p if p(x) evaluated at 5 is 48.
Since we know that 1,2,3 are roots, we can write that:
$\displaystyle P(x)=a(x-1)(x-2)(x-3)$
$\displaystyle P(5)=a(5-1)(5-2)(5-3)=24a=48$ -> $\displaystyle a=2$.
Now distribute 2 in $\displaystyle P(x)=a(x-1)(x-2)(x-3)$:
$\displaystyle P(x)=2(x-1)(x-2)(x-3)=(2x-2)(x-2)(x-3)=$
$\displaystyle =(2x^2-6x+4)(x-3)=2x^3-12x^2+22x-12$.
So $\displaystyle P(x)=2x^3-12x^2+22x-12$.
I hope you understand now.