Question 691019
Since the graph passes through the points (-3,0),(-2,0), and (1,0),
your cubic function has zeros at {{{x=-3}}}, {{{x=-2}}}, and {{{x=1}}}.
If a polynomial has a zero at {{{x=a}}}, it has (x-a) as a factor.
So your cubic function has {{{(x+3)(x+2)(x-1)=x^3+ax^2+x-6}}} as a factor.
Since that is already a cubic, the only other factor we need is a real number {{{b}}}
Your cubic function is
{{{f(x)=b(x+3)(x+2)(x-1)=b(x^3+ax^2+x-6)}}}
For {{{x=0}}}, we know that f(0)=-6, because the graph passes through (0,-6).
Then, {{{f(0)=b(-6)=-6}}} --> {{{b=1}}} and
{{{highlight(f(x)=(x+3)(x+2)(x-1)=x^3+ax^2+x-6)}}}