document.write( "Question 1103056: Factorise \"x%5E2-2x-3\" \n" ); document.write( "
Algebra.Com's Answer #717746 by richwmiller(17219)\"\" \"About 
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Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor \"1%2Ax%5E2%2B-2%2Ax%2B-3\", first we need to ask ourselves: What two numbers multiply to -3 and add to -2? Lets find out by listing all of the possible factors of -3
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\n" ); document.write( " Factors:
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\n" ); document.write( " 1,3,
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\n" ); document.write( " -1,-3,List the negative factors as well. This will allow us to find all possible combinations
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\n" ); document.write( " These factors pair up to multiply to -3.
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\n" ); document.write( " (-1)*(3)=-3
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\n" ); 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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First Number|Second Number|Sum
1|-3|1+(-3)=-2
-1|3|(-1)+3=2
We can see from the table that 1 and -3 add to -2.So the two numbers that multiply to -3 and add to -2 are: 1 and -3\r\n" ); document.write( " \r\n" ); 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( " \"%28x%2Ba%29%28x%2Bb%29\"substitute a=1 and b=-3\r\n" ); document.write( " \r\n" ); document.write( " So the equation becomes:\r\n" ); document.write( " \r\n" ); document.write( " (x+1)(x-3)\r\n" ); document.write( " \r\n" ); document.write( " Notice that if we foil (x+1)(x-3) we get the quadratic \"1%2Ax%5E2%2B-2%2Ax%2B-3\" again\n" ); document.write( "
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