document.write( "Question 788072: factor the polynomials using whatever strategy seems appropriate. State what methods you will use and then demonstrate the methods on your problems, explaining the process as you go. Discuss any particular challenges those particular polynomials posed for the factoring.
\n" ); document.write( "6w^4-54W^2 these are two separate problems 1-XY-20^2Y^2 \n" ); document.write( "

Algebra.Com's Answer #478404 by KMST(5328) $\"\"$ $\"About$

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$\"6w%5E4-54w%5E2\"$
\n" ); document.write( "The coefficients are both multiples of $\"6\"$ , because $\"-54=6%2A%28-9%29\"$ , so we could take out $\"6\"$ as a common factor.
\n" ); document.write( "Both terms have powers of $\"w\"$ , and the smallest exponent is $\"2\"$ , and we can write $\"w%5E4\"$ as $\"w%5E2%2Aw%5E2=w%5E4\"$ , so we could take out $\"w%5E2\"$ as a common factor.
\n" ); document.write( "We can take out both actors at the same time, as $\"6w%5E2\"$
\n" ); document.write( " $\"6w%5E4=6w%5E2%2Aw%5E2\"$ and $\"-54w%5E2=6W%5E2%2A%28-9%29\"$
\n" ); document.write( " $\"6w%5E4-54w%5E2=6w%5E2%28w%5E2-9%29\"$
\n" ); document.write( "Next, we realize that $\"w%5E2-9\"$ is a difference of squares, a special product of the form
\n" ); document.write( " $\"a%5E2-b%5E2=%28a%2Bb%29%28a-b%29\"$ .
\n" ); document.write( "Specifically, $\"w%5E2-9=%28w%2B3%29%28w-3%29\"$ .
\n" ); document.write( "So the complete factorization is
\n" ); document.write( " $\"6w%5E4-54w%5E2=6w%5E2%28w%5E2-9%29=highlight%286w%5E2%28w%2B3%29%28w-3%29%29\"$ \n" ); document.write( "

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