SOLUTION: 4x^3-36x=0 4x(x^2-9)=0 4x(x-3)(x+3) Is this finished?

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Question 190927This question is from textbook saxon algebra 2
: 4x^3-36x=0
4x(x^2-9)=0
4x(x-3)(x+3)
Is this finished? This question is from textbook saxon algebra 2

Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You should have 4x%28x-3%29%28x%2B3%29=0. Now set each factor equal to zero:

4x=0, x-3=0, or x%2B3=0


and solve for "x" to get


x=0, x=3 or x=-3

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
4x^3-36x=0
4x(x^2-9)=0
4x(x-3)(x+3)=0
Is this finished?
-----------------------
Not yet.
Solve for "x":
4x = 0 or x-3 = 0 or x+3 = 0
x = 0 or x = 3 or x = -3
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Cheers,
Stan H.