The zeros are -8, -1, 2, so you can write the function as f(x)=(x-(-8))(x-(-1))(x-2)=(x+8)(x+1)(x-2).
Expand:
[tex](x+8)(x+1)(x-2)= \\
(x^2+x+8x+8)(x-2)= \\
(x^2+9x+8)(x-2)= \\
x^3-2x^2+9x^2-18x+8x-16= \\
x^3+7x^2-10x-16[/tex]
The function is f(x)=x³+7x²-10x-16.
The missing value is 7.
