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Answer:

ANSWER

x3−5x2−2x+24=0

(x+2)x+2x3−5x2−2x+24

(x+2)(x2−7x+12)

(x+2)(x2−4x−3x−12)

(x+2)(x(x−4)−3(x−4))

(x+2)(x−4)(x−3)

x=−2,3,4

Given, products of 2 zeros is 12⇒(3×4)

Therefore the required solution is 3 and 4.

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