Question

Construct a polynomial that has zeros at -2;-1; 3; third degree; P(2) = 24.

Answer 1

let the equation is
p(x)=ax^3+bx^2+cx+d
each root will satisfy this equation
p(-2)=p(-1)=p(3)=0
put x= -2 , -1 , 3
and p(2)= 24 given and solve

Answer 2

well, when you know that number `v` is zero, the root would be written as `(x-v)`
So in your case,
f(x)=(x+2)(x+1)(x-3)

now, you just need to expand to receive the desired function.