SOLUTION: 2x^4=16x I have this so far, but don't know how to finish this problem 2x^4-16x = 0 2x(x^3-8) = 0 Please explain how I finish

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 2x^4=16x I have this so far, but don't know how to finish this problem 2x^4-16x = 0 2x(x^3-8) = 0 Please explain how I finish      Log On


   

Question 162371This question is from textbook College Algebra an early functions approach
: 2x^4=16x
I have this so far, but don't know how to finish this problem
2x^4-16x = 0
2x(x^3-8) = 0
Please explain how I finish This question is from textbook College Algebra an early functions approach

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
2x%5E4+=+16x
2x%5E4+-+16x+=+0
2x%2A%28x%5E3+-+8%29+=+0
This is just like saying a%2Ab+=+0
Either a, b, or both have to be
0 in order for this to be true, so either
2x+=+0 and
x+=+0
or
x%5E3+-+8+=+0
x%5E3+=+8
x+=+2
The answer is x+=+0 and x+=+2