SOLUTION: solve please {{{x^3+3x^2-4x=0}}}

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Question 534886: solve please x%5E3%2B3x%5E2-4x=0
Found 2 solutions by Earlsdon, josmiceli:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
x%5E3%2B3x%5E2-4x+=+0 Since this is a cubic (3rd order) equation, you can expect to find 3 solutions.
x%5E3%2B3x%5E2-4x+=+0 First factor an x.
x%28x%5E2%2B3x-4%29+=+0 Apply the zero product rule.
x+=+0 or x%5E2%2B3x-4+=+0 Now factor the trinomial.
%28x-1%29%28x%2B4%29+=+0 Apply the zero product rule.
x-1+=+0 or x%2B4+=+0 so...
x+=+1 or x+=+-4
The three solutions are:
x+=+0
x+=+1
x+=+-4

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+x%5E3%2B3x%5E2-4x=0+
+x%2A%28+x%5E2+%2B+3x+-+4+%29+=+0+
+x%2A%28+x+-+1+%29%2A%28+x+%2B+4+%29+=+0+
the solutions are
+x+=+0+
+x+=+1+
+x+=+-4+
check:
+x%5E3%2B3x%5E2-4x=0+
+%28-4%29%5E3+%2B+3%2A%28-4%29%5E2-4%2A%28-4%29+=+0+
+-64+%2B+48+%2B+16+=+0+
+-64+%2B+64+=+0+
OK