Algebra.Com's Answer #55910 by jim_thompson5910(35256)![\"\"](\"/images/stars/stars-13.gif\")  
You can put this solution on YOUR website!
Does your problem look like this?\r
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document.write( "If this is the case, I'm afraid you don't have the answer (despite what the other tutor said. He must have quickly glanced at it).\r
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document.write( "First factor the numerator\r
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document.write( " | Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1) | \n" );
document.write( "In order to factor  , first we need to ask ourselves: What two numbers multiply to -14 and add to 5? Lets find out by listing all of the possible factors of -14
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document.write( " Factors:
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document.write( " 1,2,7,14,
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document.write( " -1,-2,-7,-14,List the negative factors as well. This will allow us to find all possible combinations
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document.write( " These factors pair up to multiply to -14.
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document.write( " (-1)*(14)=-14
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document.write( " (-2)*(7)=-14
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document.write( " Now which of these pairs add to 5? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 5
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document.write( " | First Number | | | Second Number | | | Sum | | 1 | | | -14 | || | 1+(-14)=-13 | | 2 | | | -7 | || | 2+(-7)=-5 | | -1 | | | 14 | || | (-1)+14=13 | | -2 | | | 7 | || | (-2)+7=5 | We can see from the table that -2 and 7 add to 5.So the two numbers that multiply to -14 and add to 5 are: -2 and 7\r\n" );
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document.write( " Now we substitute these numbers into a and b of the general equation of a product of linear factors which is:\r\n" );
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document.write( "  substitute a=-2 and b=7\r\n" );
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document.write( " So the equation becomes:\r\n" );
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document.write( " (x-2)(x+7)\r\n" );
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document.write( " Notice that if we foil (x-2)(x+7) we get the quadratic again\n" );
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document.write( "Factor the denominator\r
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document.write( " | Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1) | \n" );
document.write( "In order to factor  , first we need to ask ourselves: What two numbers multiply to 21 and add to 10? Lets find out by listing all of the possible factors of 21
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document.write( " Factors:
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document.write( " 1,3,7,21,
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document.write( " -1,-3,-7,-21,List the negative factors as well. This will allow us to find all possible combinations
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document.write( " These factors pair up to multiply to 21.
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document.write( " 1*21=21
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document.write( " 3*7=21
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document.write( " (-1)*(-21)=21
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document.write( " (-3)*(-7)=21
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document.write( " note: remember two negative numbers multiplied together make a positive number
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document.write( " Now which of these pairs add to 10? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 10
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document.write( " | First Number | | | Second Number | | | Sum | | 1 | | | 21 | || | 1+21=22 | | 3 | | | 7 | || | 3+7=10 | | -1 | | | -21 | || | -1+(-21)=-22 | | -3 | | | -7 | || | -3+(-7)=-10 | We can see from the table that 3 and 7 add to 10. So the two numbers that multiply to 21 and add to 10 are: 3 and 7\r\n" );
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document.write( " Now we substitute these numbers into a and b of the general equation of a product of linear factors which is:\r\n" );
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document.write( " substitute a=3 and b=7\r\n" );
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document.write( " So the equation becomes:\r\n" );
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document.write( " (x+3)(x+7)\r\n" );
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document.write( " Notice that if we foil (x+3)(x+7) we get the quadratic again\n" );
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document.write( "So when we factor the numerator and denominator we get:\r
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document.write( " Cancel like terms\r
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document.write( " So this is your answer. \r
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document.write( "In other words, reduces to \r
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document.write( "note: you were close though. You just had an extra x in your denominator.\r
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document.write( "Notice if you graph you get\r
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document.write( " graph of \r
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document.write( "it should be the same as the graph of \r
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document.write( " graph of \r
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document.write( "We can see that they are equal. So this verifies our answer. \n" );
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