document.write( "Question 75090: how can i factor this completely? 9x^2+12xy+4y^2-25 \n" ); document.write( "
Algebra.Com's Answer #53947 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
You could try this:
\n" ); document.write( "\"9x%5E2%2B12xy%2B4y%5E2-25\" Try grouping the terms as follows:
\n" ); document.write( "\"%289x%5E2%2B12xy%2B4y%5E2%29-%2825%29%29\" Do you notice that, in the first group, the first and last terms are perfect squares> This suggests that this trinomial might be a perfect square. Let's try it.
\n" ); document.write( "\"9x%5E2%2B12xy%2B4y%5E2+=+%283x%2B2y%29%283x%2B2y%29\" = \"%283x%2B2y%29%5E2\" Yes...a perfect square! So we can rewrite the original expression as:
\n" ); document.write( "\"%289x%5E2%2B12xy%2B4y%5E2%29-%2825%29+=+%283x%2B2y%29%5E2-%285%29%5E2\" Now you can see that we have the difference of two squares which can be factored thusly: \"A%5E2-B%5E2+=+%28A%2BB%29%28A-B%29\" Applying this to your expression, we get:
\n" ); document.write( "\"%283x%2B2y%29%5E2-%285%29%5E2+=+%28%283x%2B2y%29%2B5%29%28%283x%2B2y%29-5%29\"
\n" ); document.write( "The final answer looks like:
\n" ); document.write( "\"9x%5E2%2B12xy%2B4y%5E2-25+=+%283x%2B2y%2B5%29%283x%2B2y-5%29\"
\n" ); document.write( "Check: Using the FOIL method, we'll multiply the factors to see if we get the original expression back.
\n" ); document.write( "\"%283x%2B2y%2B5%29%283x%2B2y-5%29+=+9x%5E2%2B6xy-15x%2B6xy%2B4y%5E2-10y%2B15x%2B10y-25\" Simplifying this, we get:\"9x%5E2%2B12xy%2B4y%5E2-25\" the original expression.
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