SOLUTION: This is a problem off of the practice equation problems from this site. I started to work out this problem: 2x^4-54x. I got as far as 2x(x^3-27). I couldn't figure out how they mov

Algebra ->  Equations -> SOLUTION: This is a problem off of the practice equation problems from this site. I started to work out this problem: 2x^4-54x. I got as far as 2x(x^3-27). I couldn't figure out how they mov      Log On


   

Question 66007: This is a problem off of the practice equation problems from this site. I started to work out this problem: 2x^4-54x. I got as far as 2x(x^3-27). I couldn't figure out how they moved from that to the answer of 2x(x-3)(x^2+3x+9)
Can you show the steps on how to get that final answer? I tried to figure it out on my own, but I wasn't even close. Thanks!
Found 2 solutions by Cintchr, Earlsdon:
Answer by Cintchr(481) About Me  (Show Source):
You can put this solution on YOUR website!
+2x%5E4-54x+
factor out a 2x
+2x%28x%5E3-27%29+
look at the portion in the paranthesis.
+x%5E3+-+27+
This is the difference of two cubes.
There are 2 different "special patterns" to memorize here.
One for the sum of two cubes, the other for the difference of two cubes.
1. +X%5E3%2B+Y%5E3+=+%28x+%2B+y%29%28X%5E2+-+XY+%2B+Y%5E2%29+
2. +X%5E3+-Y%5E3+=+%28X+-+Y%29%28X%5E2+%2B+XY+%2B+Y%5E2%29+
We are using the second formula/pattern
+2x%28x%5E3-27%29+
the cube root of x^3 = x
the cube root of 27 = 3
+2x+%28x+-+3%29%28x%5E2%2B3x%2B9%29+

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Factor:
2x%5E4-54x Factor the 2x as you did.
2x%28x%5E3-27%29 Now you may recognise the parentheses as the difference of two cubes for which the factored form is: A%5E3+-+B%5E3+=+%28A-B%29%28A%5E2%2BAB%2BB%5E2%29
Apply this to your equation: A = x and B = 3
2x%28x%5E3-27%29+=+2x%28%28x%29%5E3+-+%283%29%5E3%29 =