Algebra.Com's Answer #56558 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( "In order to simplify this algebraic expression, we need to factor each term\r
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document.write( "So lets factor the first 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 -8 and add to -2? Lets find out by listing all of the possible factors of -8
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document.write( " Factors:
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document.write( " 1,2,4,8,
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document.write( " -1,-2,-4,-8,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 -8.
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document.write( " (-1)*(8)=-8
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document.write( " (-2)*(4)=-8
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document.write( " Now which of these pairs add to -2? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -2
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document.write( " | First Number | | | Second Number | | | Sum | | 1 | | | -8 | || | 1+(-8)=-7 | | 2 | | | -4 | || | 2+(-4)=-2 | | -1 | | | 8 | || | (-1)+8=7 | | -2 | | | 4 | || | (-2)+4=2 | We can see from the table that 2 and -4 add to -2.So the two numbers that multiply to -8 and add to -2 are: 2 and -4\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=-4\r\n" );
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document.write( " So the equation becomes:\r\n" );
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document.write( " (x+2)(x-4)\r\n" );
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document.write( " Notice that if we foil (x+2)(x-4) we get the quadratic again\n" );
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document.write( "Now lets factor the 1st denominator  \r
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document.write( "Notice this is a difference of squares so we have \r
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document.write( "Now lets factor the 2nd 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 6 and add to -5? Lets find out by listing all of the possible factors of 6
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document.write( " Factors:
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document.write( " 1,2,3,6,
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document.write( " -1,-2,-3,-6,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 6.
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document.write( " 1*6=6
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document.write( " 2*3=6
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document.write( " (-1)*(-6)=6
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document.write( " (-2)*(-3)=6
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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 -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 | | | 6 | || | 1+6=7 | | 2 | | | 3 | || | 2+3=5 | | -1 | | | -6 | || | -1+(-6)=-7 | | -2 | | | -3 | || | -2+(-3)=-5 | We can see from the table that -2 and -3 add to -5.So the two numbers that multiply to 6 and add to -5 are: -2 and -3\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=-3\r\n" );
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document.write( " So the equation becomes:\r\n" );
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document.write( " (x-2)(x-3)\r\n" );
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document.write( " Notice that if we foil (x-2)(x-3) we get the quadratic again\n" );
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document.write( "Now lets factor the 2nd 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 -12 and add to -1? Lets find out by listing all of the possible factors of -12
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document.write( " Factors:
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document.write( " 1,2,3,4,6,12,
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document.write( " -1,-2,-3,-4,-6,-12,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 -12.
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document.write( " (-1)*(12)=-12
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document.write( " (-2)*(6)=-12
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document.write( " (-3)*(4)=-12
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document.write( " Now which of these pairs add to -1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -1
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document.write( " | First Number | | | Second Number | | | Sum | | 1 | | | -12 | || | 1+(-12)=-11 | | 2 | | | -6 | || | 2+(-6)=-4 | | 3 | | | -4 | || | 3+(-4)=-1 | | -1 | | | 12 | || | (-1)+12=11 | | -2 | | | 6 | || | (-2)+6=4 | | -3 | | | 4 | || | (-3)+4=1 | We can see from the table that 3 and -4 add to -1.So the two numbers that multiply to -12 and add to -1 are: 3 and -4\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=-4\r\n" );
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document.write( " So the equation becomes:\r\n" );
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document.write( " (x+3)(x-4)\r\n" );
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document.write( " Notice that if we foil (x+3)(x-4) we get the quadratic again\n" );
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document.write( "So our whole expression factors to\r
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document.write( " Notice these terms cancel\r
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document.write( "So we're left with\r
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document.write( "Now factor out a -1 to make t-2 become 2-t\r
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document.write( " Notice these terms cancel\r
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document.write( "Answer:
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document.write( "So the whole thing reduces to \r
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