Find a degree 3 polynomial that has zeros - 1, 3 and 8 and in which the coefficient of x^2 is - 40

Answer is 4x^3-40x^2+52x+96

Step-by-step explanation:

(x+1) (x-3) (x-8) =

(x^2-3x+x-3) (x-8) =

(x^2 - 2x-3) (x-8) =

x^3-8x^2-2x^2+16x-3x+24=

x^3-10x^2+13x+24

because coeficient of x^2 is - 40, we multiply the whole equation by 4, then

4x^3-40x^2+52x+96.