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# solve the equation y^3+2y^2=9y+18

y3 + 2y2 = 9y + 18

First, set the equation equal to 0 by subtracting the right-hand side from both sides of the equation:

y3 + 2y2 - 9y - 18 = 0

We solve this polynomial equation by factoring. To do so, consider the first two terms and last two terms separately and look to see if they contain a greatest common factor that you can factor out of each set separately:

y3 + 2y2   ==>   the gcf here is 'y2'   ==>   y2(y + 2)

-9y - 18   ==>   the gcf here is '-9'   ==>   -9(y + 2)

Thus,

y3 + 2y2 - 9y - 18 = 0

[y2(y + 2)] + [-9(y + 2)] = 0

Notice that there is now a greatest common factor among these two terms, that being 'y + 2'. So after factoring out this gcf, we arrive at the following:

(y + 2)(y2 - 9) = 0

By the zero product property, we can now set each binomial equal to 0 and solve for y:

y + 2 = 0    ==>    subtract 2 from both sides of the equation

y = -2

y- 9 = 0    ==>    add 9 to both sides of the equation

y2 = 9    ==>    take the square root of both sides of the equation

√y2 = ±√9

y = ±3

Thus, there are 3 solutions for this equation. Those being,  y = -2 ,  y = -3 ,  and  y = 3.