x^3-2x^2-4x+8 zeros? multiplicity?

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Answer 1

x^3-2x^2-4x+8 zeros? multiplicity? to find the zeros set y = 0 to get the x-axis zeros. set x = 0 to get the y-axis zeros. x^3 - 2x^2 - 4x + 8 = 0 x^3 - 2x^2 - 4x = -8 x (x^2 - 2x -4) = -8 x = -8 the first factor. need to use quadratic formula to go any further: -8 =( -(-2) + sqrt((-2)^2 +4*1*-4)))/2 and -8 = (-(-2) - sqrt((-2)^2 +4*1*-4)))/2 -8 = 2 + sqrt(4 -16)/2 and -8 = 2 + sqrt(4 - 16)/2 these root are imaginary. the only real root is -8. there are no duplicate or multiplicity in this.