document.write( "Question 701163: s2-10st+25t2.
\n" ); document.write( "m2-16n2.
\n" ); document.write( "5x2-125y2. \n" ); document.write( "
Algebra.Com's Answer #432287 by Simnepi(216)\"\" \"About 
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\"s%5E2-10st%2B25t%5E2\" from the binomial theorem we know that \"%28a-b%29%5E2=a%5E2-2ab%2Bb%5E2\"
\n" ); document.write( "Comparing the equations we see that \"a%5E2=s%5E2\" therefore a=s (by taking the square root of both sides)
\n" ); document.write( "Also we see that
\n" ); document.write( "\"b%5E2=25t%5E2\" from this we deduce that b=5t .
\n" ); document.write( "giving \"s%5E2-10st%2B25t%5E2=%28s-5t%29%5E2\"
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\n" ); document.write( "\"m%5E2-16n%5E2\" again, using the binomial theorem we know that \"%28a%2Bb%29%28a-b%29=a%5E2-b%5E2\"
\n" ); document.write( "Comparing the equations we see that \"a%5E2=m%5E2\" therefore a=m.
\n" ); document.write( "Also we see that
\n" ); document.write( "\"b%5E2=16n%5E2\" again, we can take the square root of both sides to find b=4n
\n" ); document.write( "Giving \"m%5E2-16n%5E2=%28m%2B4n%29%28m-4n%29\"
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\n" ); document.write( "\"5x%5E2-125y%5E2\" let's look for a common factor! Both terms can be divided by 5 so lets take that out.
\n" ); document.write( "\"5%28x%5E2-25y%5E2%29\"Now this looking like in the second problem.
\n" ); document.write( "using the binomial theorem we know that \"%28a%2Bb%29%28a-b%29=a%5E2-b%5E2\"
\n" ); document.write( "Comparing the equations \"a%5E2=x%5E2\" giving a=x
\n" ); document.write( "and \"b%5E2=25y%5E2\" from which we deduce b=5y
\n" ); document.write( "This gives
\n" ); document.write( "\"5x%5E2-125y%5E2=5%28x%2B5y%29%28x-5y%29\"
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