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Question 408103: when the cube of a number is added to twice its square, the result is equal to 18 more than 9 times the number
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! when the cube of a number is added to twice its square,
the result is equal to 18 more than 9 times the number
:
x^3 + 2x^2 = 9x + 18
x^3 + 2x^2 - 9x - 18 = 0
Factor by grouping
x^2(x + 2) - 9(x + 2)
Factor out (x+2) and you have:
(x^2 - 9)(x + 2) = 0
x = -2
and
x^2 - 9 = 0
difference of squares
(x-3)(x+3) = 0
x = 3
x = -3
:
Check all three solutions in original equation
x^3 + 2x^2 = 9x + 18
x=-2
(-2)^3 + 2(-2)^2 = 9(-2) + 18
-8 + 8 = -18 + 18; a good solution
:
x=3
3^3 + 2(3^2) = 9(3) + 18
27 + 18 = 27 + 18; a good solution
:
(-3)^3 + 2(-3^2) = 9(-3) + 18
-27 + 18 = -27 + 18; a good solution also
:
we can say, x = -2, -3, +3, are all solutions of x
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