document.write( "Question 723112: How do you factor this completely? 6rcubed+15rsquared+8r+20 \n" ); document.write( "

Algebra.Com's Answer #443196 by KMST(5328) $\"\"$ $\"About$

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The polynomial $\"P%28r%29=6r%5E3%2B15r%5E2%2B8r%2B20\"$ would have a factor of the form $\"%28qr%2Bp%29\"$
\n" ); document.write( "if $\"-p%2Fq\"$ is a zero of the polynomial.
\n" ); document.write( "Zeros of that polynomial must be negative because for all $\"r%3E=0\"$ , $\"6r%5E3%2B15r%5E2%2B8r%3E=0\"$ and $\"6r%5E3%2B15r%5E2%2B8r%2B20%3E=20\"$
\n" ); document.write( "Any rational zeros will be of the form $\"-p%2Fq\"$ ,
\n" ); document.write( "where $\"p\"$ is a factor of the independent term, $\"20\"$ , and
\n" ); document.write( " $\"q\"$ is a factor of the leading coefficient $\"6\"$ .
\n" ); document.write( "So, $\"q\"$ could be 1, 2, 3, or 6, and $\"p\"$ could be 1, 2, 4, 5, 10, or 20.
\n" ); document.write( " $\"P%28-1%29=-6%2B15-8%2B20=21\"$
\n" ); document.write( " $\"P%28-2%29=-8%2A6%2B15%2A4-2%2A8%2B20=-48%2B60-16%2B20=16\"$
\n" ); document.write( " $\"P%28-4%29=-64%2A6%2B15%2A16-4%2A8%2B20=-286%2B90-32%2B20=-58\"$
\n" ); document.write( "Since $\"P%28-2%29%3E0\"$ and $\"P%28-4%29%3C0\"$ , there must be a zero of $\"P%28r%29\"$ between -2 and -4. Of all the rational possibilities, only $\"-5%2F2\"$ and $\"-10%2F3\"$ are between -2 and -4.
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\n" ); document.write( "so $\"%282r%2B5%29\"$ is a factor of $\"P%28r%29\"$ and dividing $\"P%28r%29\"$ by $\"%282r%2B5%29\"$ we get $\"3r%5E2%2B4\"$ , which cannot be factored further, so
\n" ); document.write( " $\"6r%5E3%2B15r%5E2%2B8r%2B20=highlight%28%282r%2B5%29%283r%5E2%2B4%29%29\"$
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\n" ); document.write( "NOTE: If you have gone beyond real numbers and are expected to use complex numbers, you could factor $\"%283r%5E2%2B4%29\"$ further. \n" ); document.write( "

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