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Verified Answer :

apologiabiology

a polynomial with roots/zeros at r1,r2,r3 has factors
(x-r1)(x-r2)(x-r3)

zeroes -6,5,-2
(x-(-6))(x-5)(x-(-2)) or
(x+6)(x-5)(x+2)

the function is f(x)=(x+6)(x+2)(x-5) or expanded f(x)=x^3+3x^2-28x-60

Lampa20

given zeroes are:
x1 = -6
x2 = 5
x3 = -2

The general formula for cubic function that has zeroes is: (factor form)
(x-x1)(x-x2)(x-x3)

if we express our zeroes we get:
(x+6)(x-5)(x+2) = (x^2 - 5x +6x - 30)(x+2) = (x^2 + x - 30) ( x+2)=
x^3 + x^2 -30x +2x^2 + 2x - 60 = x^3 +3x^2 -32x - 60

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