finding cubic function using algebraic method?

Question

i need help on how to find the cubic function using algebraic method with a given table of values. No calculator. (-2,0) (-1,24) (1,0) (5,0)

p.s. if u have time, can u answer my survey question.

answer is not (x+2)(x-1)(x-5)

Answer 1

the cubic function
f(x)=ax^3+bx^2+cx+d =a[x^3+ (b/a)x^2+(c/a)x+(d/a)]
if f(x)=0 (-2,0)(1,0)(5,0)
b/a= -(r1+r2+r3)=-(-2+1+5)= -4 (r1,r2,r3 are the roots)
c/a=(r1r2+r1r3+r2r3) = (-2)*1+(-2)*5+1*5=-2-10+5= -7
d/a=-(r1r2r3)= -( -2*1*5 )= 10
so f(x)=a[x^3 - 4 x^2 -7 x +10]
to get 'a' use the 4th point (-1,24)
f(-1)=24=a[-1-4+7+10]=12a
a= 2
so f(x)= 2 x^3- 8 x^2 -14 x + 20

Answer 2

The function is y = (x+2)(x-1)(x-5)

Answer 3

I got a better Q. If I answer your questions ten time, and I get two points, how long will it take me to get to lever 6?