In order to factor  , first we need to ask ourselves: What two numbers multiply to -15 and add to 2? Lets find out by listing all of the possible factors of -15
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document.write( " Factors:
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document.write( " 1,3,5,15,
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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( " 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 | | | -15 | || | 1+(-15)=-14 | | 3 | | | -5 | || | 3+(-5)=-2 | | -1 | | | 15 | || | (-1)+15=14 | | -3 | | | 5 | || | (-3)+5=2 | We can see from the table that -3 and 5 add to 2.So the two numbers that multiply to -15 and add to 2 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" );
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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( " Notice that if we foil (x-3)(x+5) we get the quadratic again\n" );
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document.write( "Factor the bottom 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 -25 and add to 0? Lets find out by listing all of the possible factors of -25
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document.write( " Factors:
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document.write( " 1,5,25,15,
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document.write( " -1,-5,-25,-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 -25.
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document.write( " (-1)*(15)=-25
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document.write( " (-5)*(25)=-25
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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 | | | -15 | || | 1+(-15)=-14 | | 5 | | | -25 | || | 5+(-25)=-20 | | -1 | | | 15 | || | (-1)+15=14 | | -5 | | | 25 | || | (-5)+25=20 | 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( " Notice that if we foil (x-3)(x+5) we get the quadratic again\r\n" );
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document.write( " None of these factors add to 0. So this quadratic cannot be factored. In order to solve for x, we need to use the quadratic formula.\n" );
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document.write( "So now we have\r
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document.write( " Factor a 2 out of  \r
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document.write( " Flip the 2nd fraction and multiply\r
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document.write( " Cancel like terms\r
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document.write( " Reduce. This is the simplified form. \n" );
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