Question 1031408
Notice that the points (-5,0), (2,0), (3,0) give the x-intercepts of the cubic function, hence the roots of that function.

==> {{{f(x) = k(x+5)(x-2)(x-3) }}}

for some undetermined constant k.

Since f(0) = 20, ==> {{{f(0) = k(0+5)(0-2)(0-3) = 20 }}} ==> {{{k = 2/3}}}.

Therefore the function is {{{f(x) = (2/3)(x+5)(x-2)(x-3) }}}, which is just 

{{{f(x) = (2/3)x^3 - (38/3)x + 20}}}.