Respuesta :

Given f(x)=x^3+8x^2-4x-32 is a cubic eqn w/ 3 distinct zeros: -8, -2 n 2

the multiplicity for each zero will be 1...as the eqn is in cubic order.
f(x)=x^3+8x^2-4x-32
note that f(2)=0 so x=2 is one of the zero
dividing f(x) by (x-2)
the quotient is x^2+10x+16 which factors into (x+2)(x+8)
combining f(x)=(x-2)(x+2)(x+8)

so the zeroes are -8, -2 and +2 and the multiplicity is 1 for all zeroes

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