document.write( "Question 708860: y^3+8y^2-y-8 \n" ); document.write( "

Algebra.Com's Answer #436290 by jim_thompson5910(35256) $\"\"$ $\"About$

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$\"y%5E3%2B8y%5E2-y-8\"$ Start with the given expression\r
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\n" ); document.write( "\n" ); document.write( " $\"%28y%5E3%2B8y%5E2%29%2B%28-y-8%29\"$ Group the terms in two pairs\r
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\n" ); document.write( "\n" ); document.write( " $\"y%5E2%28y%2B8%29-1%28y%2B8%29\"$ Factor out the GCF $\"y%5E2\"$ out of the first group. Factor out the GCF $\"-1\"$ out of the second group\r
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\n" ); document.write( "\n" ); document.write( " $\"%28y%5E2-1%29%28y%2B8%29\"$ Since we have the common term $\"y%2B8\"$ , we can combine like terms\r
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\n" ); document.write( "\n" ); document.write( " $\"%28y-1%29%28y%2B1%29%28y%2B8%29\"$ Factor $\"y%5E2-1\"$ to get $\"%28y-1%29%28y%2B1%29\"$ using the difference of squares rule\r
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\n" ); document.write( "\n" ); document.write( "So $\"y%5E3%2B8y%5E2-y-8\"$ factors to $\"%28y-1%29%28y%2B1%29%28y%2B8%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "In other words, $\"y%5E3%2B8y%5E2-y-8=%28y-1%29%28y%2B1%29%28y%2B8%29\"$ for all values of y
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