SOLUTION: factor by grouping 5x^2-12x+7

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Question 191090: factor by grouping
5x^2-12x+7

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

Looking at the expression 5x%5E2-12x%2B7, we can see that the first coefficient is 5, the second coefficient is -12, and the last term is 7.


Now multiply the first coefficient 5 by the last term 7 to get %285%29%287%29=35.


Now the question is: what two whole numbers multiply to 35 (the previous product) and add to the second coefficient -12?


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 -12:


First NumberSecond NumberSum
1351+35=36
575+7=12
-1-35-1+(-35)=-36
-5-7-5+(-7)=-12



From the table, we can see that the two numbers -5 and -7 add to -12 (the middle coefficient).


So the two numbers -5 and -7 both multiply to 35 and add to -12


Now replace the middle term -12x with -5x-7x. Remember, -5 and -7 add to -12. So this shows us that -5x-7x=-12x.


5x%5E2%2Bhighlight%28-5x-7x%29%2B7 Replace the second term -12x with -5x-7x.


%285x%5E2-5x%29%2B%28-7x%2B7%29 Group the terms into two pairs.


5x%28x-1%29%2B%28-7x%2B7%29 Factor out the GCF 5x from the first group.


5x%28x-1%29-7%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.


%285x-7%29%28x-1%29 Combine like terms. Or factor out the common term x-1

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Answer:


So 5x%5E2-12x%2B7 factors to %285x-7%29%28x-1%29.


Note: you can check the answer by FOILing %285x-7%29%28x-1%29 to get 5x%5E2-12x%2B7 or by graphing the original expression and the answer (the two graphs should be identical).