SOLUTION: Hi, I needed help on this question...I don't know how to do it. Find the solutions: 2x to the 4th + 8x to the 3rd = 0

Question 738152: Hi, I needed help on this question...I don't know how to do it.
Find the solutions:
2x to the 4th + 8x to the 3rd = 0
Found 2 solutions by richwmiller, DrBeeee:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
2x^4+8x^3=0
2x^3*(x+4)=0
x=0
x=-4

Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
I'm not sure you mean the given equation is
(1) (2x)^4 + (8x)^3 = 0 OR
(2) 2x^4 + 8x^3 = 0
It doesn't effect the solution technicque, only the value of one of the roots.
I'll do (1), where we have
(3) 16x^4 + 512x^3 = 0
Dividing by 16 gives us
(4) x^4 + 32x^3 = 0
Now factor out x^3 and get
(5) x^3*(x + 32) = 0
Since the given equation is 4th order, we need four roots,. They are given by the solution set x = {0,0,0,-32). Note that x^3 = 0 results in a triple root at x = 0.
If you meant that the given equation is given by (2), the solution set is {0,0,0,-4)
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