document.write( "Question 93008: x^2-x-20 \r
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Algebra.Com's Answer #67689 by jim_thompson5910(35256)\"\" \"About 
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression \"x%5E2-x-20\", we can see that the first coefficient is \"1\", the second coefficient is \"-1\", and the last term is \"-20\".



Now multiply the first coefficient \"1\" by the last term \"-20\" to get \"%281%29%28-20%29=-20\".



Now the question is: what two whole numbers multiply to \"-20\" (the previous product) and add to the second coefficient \"-1\"?



To find these two numbers, we need to list all of the factors of \"-20\" (the previous product).



Factors of \"-20\":

1,2,4,5,10,20

-1,-2,-4,-5,-10,-20



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to \"-20\".

1*(-20) = -20
2*(-10) = -20
4*(-5) = -20
(-1)*(20) = -20
(-2)*(10) = -20
(-4)*(5) = -20


Now let's add up each pair of factors to see if one pair adds to the middle coefficient \"-1\":



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First NumberSecond NumberSum
1-201+(-20)=-19
2-102+(-10)=-8
4-54+(-5)=-1
-120-1+20=19
-210-2+10=8
-45-4+5=1




From the table, we can see that the two numbers \"4\" and \"-5\" add to \"-1\" (the middle coefficient).



So the two numbers \"4\" and \"-5\" both multiply to \"-20\" and add to \"-1\"



Now replace the middle term \"-1x\" with \"4x-5x\". Remember, \"4\" and \"-5\" add to \"-1\". So this shows us that \"4x-5x=-1x\".



\"x%5E2%2Bhighlight%284x-5x%29-20\" Replace the second term \"-1x\" with \"4x-5x\".



\"%28x%5E2%2B4x%29%2B%28-5x-20%29\" Group the terms into two pairs.



\"x%28x%2B4%29%2B%28-5x-20%29\" Factor out the GCF \"x\" from the first group.



\"x%28x%2B4%29-5%28x%2B4%29\" Factor out \"5\" from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



\"%28x-5%29%28x%2B4%29\" Combine like terms. Or factor out the common term \"x%2B4\"



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Answer:



So \"x%5E2-x-20\" factors to \"%28x-5%29%28x%2B4%29\".



In other words, \"x%5E2-x-20=%28x-5%29%28x%2B4%29\".



Note: you can check the answer by expanding \"%28x-5%29%28x%2B4%29\" to get \"x%5E2-x-20\" or by graphing the original expression and the answer (the two graphs should be identical).

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