In order to factor  , first we need to ask ourselves: What two numbers multiply to -9 and add to 0? Lets find out by listing all of the possible factors of -9
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document.write( " Factors:
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document.write( " 1,3,9,
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document.write( " -1,-3,-9,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 -9.
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document.write( " (-1)*(9)=-9
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document.write( " (-3)*(3)=-9
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document.write( " Now which of these pairs add to 0? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 0
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document.write( " | First Number | | | Second Number | | | Sum | | 1 | | | -9 | || | 1+(-9)=-8 | | 3 | | | -3 | || | 3+(-3)=0 | | -1 | | | 9 | || | (-1)+9=8 | | -3 | | | 3 | || | (-3)+3=0 | We can see from the table that -3 and 3 add to 0.So the two numbers that multiply to -9 and add to 0 are: -3 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=-3 and b=3\r\n" );
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document.write( " So the equation becomes:\r\n" );
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document.write( " (x-3)(x+3)\r\n" );
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document.write( " Notice that if we foil (x-3)(x+3) we get the quadratic again\n" );
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document.write( "So the denominator  factors to:
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document.write( "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 10 and add to 7? Lets find out by listing all of the possible factors of 10
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document.write( " Factors:
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document.write( " 1,2,5,10,
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document.write( "
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document.write( " -1,-2,-5,-10,List the negative factors as well. This will allow us to find all possible combinations
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document.write( "
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document.write( " These factors pair up to multiply to 10.
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document.write( " 1*10=10
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document.write( " 2*5=10
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document.write( " (-1)*(-10)=10
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document.write( " (-2)*(-5)=10
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document.write( " note: remember two negative numbers multiplied together make a positive number
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document.write( "
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document.write( " Now which of these pairs add to 7? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 7
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document.write( " | First Number | | | Second Number | | | Sum | | 1 | | | 10 | || | 1+10=11 | | 2 | | | 5 | || | 2+5=7 | | -1 | | | -10 | || | -1+(-10)=-11 | | -2 | | | -5 | || | -2+(-5)=-7 | We can see from the table that 2 and 5 add to 7. So the two numbers that multiply to 10 and add to 7 are: 2 and 5\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=5\r\n" );
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document.write( " So the equation becomes:\r\n" );
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document.write( " (x+2)(x+5)\r\n" );
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document.write( " Notice that if we foil (x+2)(x+5) we get the quadratic again\n" );
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document.write( "So the numerator  factors to:
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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 15 and add to 8? Lets find out by listing all of the possible factors of 15
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document.write( "
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document.write( " Factors:
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document.write( " 1,3,5,15,
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document.write( "
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document.write( " -1,-3,-5,-15,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 15.
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document.write( " 1*15=15
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document.write( " 3*5=15
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document.write( " (-1)*(-15)=15
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document.write( " (-3)*(-5)=15
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document.write( " note: remember two negative numbers multiplied together make a positive number
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document.write( "
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document.write( " Now which of these pairs add to 8? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 8
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document.write( " | First Number | | | Second Number | | | Sum | | 1 | | | 15 | || | 1+15=16 | | 3 | | | 5 | || | 3+5=8 | | -1 | | | -15 | || | -1+(-15)=-16 | | -3 | | | -5 | || | -3+(-5)=-8 | We can see from the table that 3 and 5 add to 8. So the two numbers that multiply to 15 and add to 8 are: 3 and 5\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" );
document.write( " \r\n" );
document.write( " substitute a=3 and b=5\r\n" );
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document.write( " So the equation becomes:\r\n" );
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document.write( " (x+3)(x+5)\r\n" );
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document.write( " \r\n" );
document.write( " Notice that if we foil (x+3)(x+5) we get the quadratic again\n" );
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document.write( "
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document.write( "So the denominator  factors to:
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document.write( "So the whole expression becomes\r
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document.write( "Which reduces to:\r
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