SOLUTION: x^2 + 6x - 7 i am suppose to factor for x using area and i am lost.

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Question 560713: x^2 + 6x - 7 i am suppose to factor for x using area and i am lost.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression x%5E2%2B6x-7, we can see that the first coefficient is 1, the second coefficient is 6, and the last term is -7.


Now multiply the first coefficient 1 by the last term -7 to get %281%29%28-7%29=-7.


Now the question is: what two whole numbers multiply to -7 (the previous product) and add to the second coefficient 6?


To find these two numbers, we need to list all of the factors of -7 (the previous product).


Factors of -7:
1,7
-1,-7


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


These factors pair up and multiply to -7.
1*(-7) = -7
(-1)*(7) = -7

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 6:


First NumberSecond NumberSum
1-71+(-7)=-6
-17-1+7=6



From the table, we can see that the two numbers -1 and 7 add to 6 (the middle coefficient).


So the two numbers -1 and 7 both multiply to -7 and add to 6


Now replace the middle term 6x with -x%2B7x. Remember, -1 and 7 add to 6. So this shows us that -x%2B7x=6x.


x%5E2%2Bhighlight%28-x%2B7x%29-7 Replace the second term 6x with -x%2B7x.


%28x%5E2-x%29%2B%287x-7%29 Group the terms into two pairs.


x%28x-1%29%2B%287x-7%29 Factor out the GCF x from the first group.


x%28x-1%29%2B7%28x-1%29 Factor out 7 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%2B7%29%28x-1%29 Combine like terms. Or factor out the common term x-1


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Answer:


So x%5E2%2B6x-7 factors to %28x%2B7%29%28x-1%29.


In other words, x%5E2%2B6x-7=%28x%2B7%29%28x-1%29.


Note: you can check the answer by expanding %28x%2B7%29%28x-1%29 to get x%5E2%2B6x-7 or by graphing the original expression and the answer (the two graphs should be identical).
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