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Solve for $y$: $y^3 - 3y^2 - 16y - 12 = 0$
in Grade 12 Maths by Bronze Status (5.9k points) | 25 views

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y= -1 gives 0 when substituted, hence y+1 is a factor

divide the function by y+1 using long division to get

(y+1)(y^2 -4y - 12)=0, then factorize

(y+1)(y+2)(y-6)=0

     y= -1 or y= -2 or y 6
by Diamond (80.1k points)

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