document.write( "Question 117697: factor the following polynomial as completely as possible\r
\n" ); document.write( "\n" ); document.write( "(((4c^4x^2-x^2-16c^4+4))) \n" ); document.write( "

Algebra.Com's Answer #85970 by Earlsdon(6294) $\"\"$ $\"About$

You can put this solution on YOUR website!
Hint: Next time, use { } instead of ( ) to express your equations.\r
\n" ); document.write( "\n" ); document.write( "Factor as completely as possible:
\n" ); document.write( " $\"4c%5E4x%5E2+-+x%5E2+-+16c%5E4+%2B+4\"$ Regroup the terms as shown:
\n" ); document.write( " $\"4c%5E4x%5E2+-+16c%5E4+-+x%5E2+%2B+4\"$ Now, group these into the two groups as shown, and don't forget the sign-change on the last term when add the parentheses.
\n" ); document.write( " $\"%284c%5E4x%5E2+-+16c%5E4%29+-+%28x%5E2+-+4%29\"$ Now factor each group. The first group has the common factor of $\"4c%5E4\"$ while the second group has no common factor.
\n" ); document.write( " $\"%284c%5E4%29%28x%5E2+-+4%29+-+%28x%5E2+-+4%29\"$ Now you can factor the common factor of $\"%28x%5E2+-+4%29\"$
\n" ); document.write( " $\"%28x%5E2+-+4%29%284c%5E4+-+1%29\"$ Notice now that these two terms are both \"difference of squares\" binomials and these can be further factored. We'll do on set at a time.
\n" ); document.write( " $\"%28x%5E2+-+4%29+=+%28x%2B2%29%28x-2%29\"$ and...
\n" ); document.write( " $\"%284c%5E4+-+1%29+=+%282c%5E2%2B1%29%282c%5E2-1%29\"$ so...putting it all together, we get:
\n" ); document.write( " $\"4c%5E4x%5E2-x%5E2-16c%5E4%2B4+=+%28x%2B2%29%28x-2%29%282c%5E2%2B1%29%282c%5E2-1%29\"$ \n" ); document.write( "

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