document.write( "Question 113019: State method and factor completely
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\n" ); document.write( "\n" ); document.write( "The method to be used is the grouping of terms but I am stuck on the factoring part because the 3q^ is not factorable if I'm not mistaken. Can you assist me on understanding this problem.
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Algebra.Com's Answer #82317 by jim_thompson5910(35256)\"\" \"About 
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\"2p-6q%2Bpq-3q%5E2\" Start with the given expression\r
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\n" ); document.write( "\n" ); document.write( "\"%282p-6q%29%2B%28pq-3q%5E2%29\" Group like terms\r
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\n" ); document.write( "\n" ); document.write( "\"2%28p-3q%29%2Bq%28p-3q%29\" Factor out the GCF of \"2\" out of the first group. Factor out the GCF of \"q\" out of the second group\r
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\n" ); document.write( "\n" ); document.write( "\"%282%2Bq%29%28p-3q%29\" Since we have a common term of \"p-3q\", we can combine like terms\r
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\n" ); document.write( "\n" ); document.write( "So \"2p-6q%2Bpq-3q%5E2\" factors to \"%282%2Bq%29%28p-3q%29\"
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