SOLUTION: I am totally lost on this problem, please explain and show the steps. Here is the problem: 2m^2n^2 - 32mn^2 + 8m^2 - 128 I thought the factor of this expression would be "2"

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: I am totally lost on this problem, please explain and show the steps. Here is the problem: 2m^2n^2 - 32mn^2 + 8m^2 - 128 I thought the factor of this expression would be "2"      Log On


   

Question 6256: I am totally lost on this problem, please explain and show the steps.
Here is the problem:
2m^2n^2 - 32mn^2 + 8m^2 - 128
I thought the factor of this expression would be "2", but I lost and I do not understand how this is the answer: 2(m + 4)(m - 4)(mn^2 + 4)
Confused
tsj
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
You did the first step right--you factored out the common factor of 2:
2m%5E2n%5E2+-+32mn%5E2+%2B+8m%5E2+-+128
2%28m%5E2n%5E2+-+16mn%5E2+%2B+4m%5E2+-+64%29

However what remains is something with 4 terms that factors by "grouping." From the first two terms, take out an n%5E2, and from the last two terms, take out a 4.
2%28n%5E2%28m%5E2+-+16m%29+%2B+4%28m%5E2+-+16%29+%29

At this point, I suspect that you copied the problem right. In order to factor by grouping, you MUST have a common factor after you do the grouping and factor as we did here. As you can see, this is NOT the case, because of the extra m in the first grouping. Let me suggest what I think the problem should have been:
2m%5E3n%5E2+-+32mn%5E2+%2B+8m%5E2+-+128
2%28m%5E3n%5E2+-+16mn%5E2+%2B+4m%5E2+-+64%29
2%28mn%5E2%28m%5E2+-+16%29+%2B+4%28m%5E2+-+16%29+%29

Now you can take out the common factor, which is %28m%5E2+-+16%29
2%28m%5E2-16%29+%28mn%5E2+%2B4%29+

Last, factor the m%5E2+-+16 which is the difference of two squares:
2%28m-4%29%28m%2B4%29+%28mn%5E2%2B4%29

R^2 at SCC