document.write( "Question 171359: Factoring Polynomials \r
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Algebra.Com's Answer #126514 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
First, we need to factor $\"3t%5E2%2B11t-20\"$ \r
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\n" ); document.write( "\n" ); document.write( "Looking at $\"3t%5E2%2B11t-20\"$ we can see that the first term is $\"3t%5E2\"$ and the last term is $\"-20\"$ where the coefficients are 3 and -20 respectively.\r
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\n" ); document.write( "\n" ); document.write( "Now multiply the first coefficient 3 and the last coefficient -20 to get -60. Now what two numbers multiply to -60 and add to the middle coefficient 11? Let's list all of the factors of -60:\r
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\n" ); document.write( "\n" ); document.write( "Factors of -60:\r
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\n" ); document.write( "\n" ); document.write( "-1,-2,-3,-4,-5,-6,-10,-12,-15,-20,-30,-60 ...List the negative factors as well. This will allow us to find all possible combinations\r
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\n" ); document.write( "\n" ); document.write( "These factors pair up and multiply to -60\r
\n" ); document.write( "\n" ); document.write( "(1)*(-60)\r
\n" ); document.write( "\n" ); document.write( "(2)*(-30)\r
\n" ); document.write( "\n" ); document.write( "(3)*(-20)\r
\n" ); document.write( "\n" ); document.write( "(4)*(-15)\r
\n" ); document.write( "\n" ); document.write( "(5)*(-12)\r
\n" ); document.write( "\n" ); document.write( "(6)*(-10)\r
\n" ); document.write( "\n" ); document.write( "(-1)*(60)\r
\n" ); document.write( "\n" ); document.write( "(-2)*(30)\r
\n" ); document.write( "\n" ); document.write( "(-3)*(20)\r
\n" ); document.write( "\n" ); document.write( "(-4)*(15)\r
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\n" ); document.write( "\n" ); document.write( "note: remember, the product of a negative and a positive number is a negative number\r
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\n" ); document.write( "\n" ); document.write( "Now which of these pairs add to 11? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 11\r
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First Number	Second Number	Sum
1	-60	1+(-60)=-59
2	-30	2+(-30)=-28
3	-20	3+(-20)=-17
4	-15	4+(-15)=-11
5	-12	5+(-12)=-7
6	-10	6+(-10)=-4
-1	60	-1+60=59
-2	30	-2+30=28
-3	20	-3+20=17
-4	15	-4+15=11
-5	12	-5+12=7
-6	10	-6+10=4

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\n" ); document.write( "\n" ); document.write( "From this list we can see that -4 and 15 add up to 11 and multiply to -60\r
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\n" ); document.write( "\n" ); document.write( "Now looking at the expression $\"3t%5E2%2B11t-20\"$ , replace $\"11t\"$ with $\"-4t%2B15t\"$ (notice $\"-4t%2B15t\"$ adds up to $\"11t\"$ . So it is equivalent to $\"11t\"$ )\r
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\n" ); document.write( "\n" ); document.write( " $\"3t%5E2%2Bhighlight%28-4t%2B15t%29%2B-20\"$ \r
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\n" ); document.write( "\n" ); document.write( "Now let's factor $\"3t%5E2-4t%2B15t-20\"$ by grouping:\r
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\n" ); document.write( "\n" ); document.write( " $\"%283t%5E2-4t%29%2B%2815t-20%29\"$ Group like terms\r
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\n" ); document.write( "\n" ); document.write( " $\"t%283t-4%29%2B5%283t-4%29\"$ Factor out the GCF of $\"t\"$ out of the first group. Factor out the GCF of $\"5\"$ out of the second group\r
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\n" ); document.write( "\n" ); document.write( " $\"%28t%2B5%29%283t-4%29\"$ Since we have a common term of $\"3t-4\"$ , we can combine like terms\r
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\n" ); document.write( "\n" ); document.write( "So $\"3t%5E2-4t%2B15t-20\"$ factors to $\"%28t%2B5%29%283t-4%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "So this also means that $\"3t%5E2%2B11t-20\"$ factors to $\"%28t%2B5%29%283t-4%29\"$ (since $\"3t%5E2%2B11t-20\"$ is equivalent to $\"3t%5E2-4t%2B15t-20\"$ )\r
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\n" ); document.write( "\n" ); document.write( "So $\"3t%5E2%2B11t-20\"$ factors to $\"%28t%2B5%29%283t-4%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "So this means that $\"3t%5E2%2B11t-20=0\"$ is the same as $\"%28t%2B5%29%283t-4%29=0\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"%28t%2B5%29%283t-4%29=0\"$ Start with the previous equation.\r
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\n" ); document.write( "\n" ); document.write( "Now set each factor equal to zero:\r
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\n" ); document.write( "\n" ); document.write( " $\"t%2B5=0\"$ or $\"3t-4=0\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"t=-5\"$ or $\"t=4%2F3\"$ Now solve for t in each case\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So the solutions are \r
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\n" ); document.write( "\n" ); document.write( " $\"t=-5\"$ or $\"t=4%2F3\"$ \n" ); document.write( "

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