SOLUTION: Factor each polynomial a^2-2a-35

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Question 181956: Factor each polynomial a^2-2a-35

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Looking at the expression a%5E2-2a-35, we can see that the first coefficient is 1, the second coefficient is -2, and the last term is -35.


Now multiply the first coefficient 1 by the last term -35 to get %281%29%28-35%29=-35.


Now the question is: what two whole numbers multiply to -35 (the previous product) and add to the second coefficient -2?


To find these two numbers, we need to list all of the factors of -35 (the previous product).


Factors of -35:
1,5,7,35
-1,-5,-7,-35


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


These factors pair up and multiply to -35.
1*(-35)
5*(-7)
(-1)*(35)
(-5)*(7)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient -2:


First NumberSecond NumberSum
1-351+(-35)=-34
5-75+(-7)=-2
-135-1+35=34
-57-5+7=2



From the table, we can see that the two numbers 5 and -7 add to -2 (the middle coefficient).


So the two numbers 5 and -7 both multiply to -35 and add to -2


Now replace the middle term -2a with 5a-7a. Remember, 5 and -7 add to -2. So this shows us that 5a-7a=-2a.


a%5E2%2Bhighlight%285a-7a%29-35 Replace the second term -2a with 5a-7a.


%28a%5E2%2B5a%29%2B%28-7a-35%29 Group the terms into two pairs.


a%28a%2B5%29%2B%28-7a-35%29 Factor out the GCF a from the first group.


a%28a%2B5%29-7%28a%2B5%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.


%28a-7%29%28a%2B5%29 Combine like terms. Or factor out the common term a%2B5

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Answer:


So a%5E2-2a-35 factors to %28a-7%29%28a%2B5%29.


Note: you can check the answer by FOILing %28a-7%29%28a%2B5%29 to get a%5E2-2a-35 or by graphing the original expression and the answer (the two graphs should be identical).