Algebra.Com's Answer #55473 by jim_thompson5910(35256)  You can put this solution on YOUR website! To find the x-intercepts, we need to factor \r \n" );
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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 \n" );
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document.write( " Factors: \n" );
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document.write( " 1,2,4,8, \n" );
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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 \n" );
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document.write( " These factors pair up to multiply to -8. \n" );
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document.write( " (-1)*(8)=-8 \n" );
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document.write( " (-2)*(4)=-8 \n" );
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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 \n" );
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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( "So the equation becomes\r \n" );
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document.write( "Now let y=0\r \n" );
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document.write( "Set each factor equal to zero:\r \n" );
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document.write( "So the solutions are:\r \n" );
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document.write( " or \r \n" );
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document.write( "which means the x-intercepts are:\r \n" );
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document.write( "(2,0) and (-4,0)\r \n" );
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document.write( " graph of  \n" );
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