document.write( "Question 190404: How do you factor $\"+c%5E2-7c=0+\"$ ? I only remember how to factor problems in the form $\"+ax%5E2+%2B+bx+%2B+c+\"$ \n" ); document.write( "

Algebra.Com's Answer #142899 by jim_thompson5910(35256) $\"\"$ $\"About$

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Remember, the distributive property states that $\"a%28b%2Bc%29=ab%2Bac\"$ . This can be reversed to get $\"ab%2Bac=a%28b%2Bc%29\"$ . In other words, we've factored out the common term \"a\".\r
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\n" ); document.write( "\n" ); document.write( "In the case of $\"+c%5E2-7c=0+\"$ , the term $\"c%5E2\"$ can be factored to get $\"+c%2Ac-7c=0+\"$ \r
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\n" ); document.write( "\n" ); document.write( "Notice how there's a common factor of \"c\". So factor this out to get $\"c%28c-7%29=0\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now set each term equal to zero and solve each equation. \n" ); document.write( "

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