Hi,
Ok so the question is: \(16sin(x)cos^3(x)-8sin(x)cos(x)\)
Now, if we factor \(8sin(x)cos(x)\), then there should be no sin(x) in either of the terms in the other bracket right?
\(8sin(x)cos(x)[2cos^2(x)-1]\).
To check, expand it again, you will get the question.
But your factorization: \(8sinxcosx(2sinxcos^2x-1)=16sin^2xcos^3x-8sinxcosx\)
So you put an extra sin(x) in the first term.
Here is a more clarification:
\(16sin(x)cos^3(x)-8sin(x)cos(x) \\ \text{Let us factor sin(x) first only} \\ sin(x)[16cos^3(x)-8cos(x)] \\ \text{Correct so far right?} \\ \text{Now, let's factor 8cos(x)} \\ sin(x)*8cos(x)[2cos^2(x)-1] \\\)
Which is how I got the factorization.