SOLUTION: Factor 7x^2+49x+42

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Question 151747: Factor 7x^2+49x+42
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

7x%5E2%2B49x%2B42 Start with the given expression


7%28x%5E2%2B7x%2B6%29 Factor out the GCF 7


Now let's focus on the inner expression x%5E2%2B7x%2B6




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Looking at the expression x%5E2%2B7x%2B6, we can see that the first coefficient is 1, the second coefficient is 7, and the last term is 6.


Now multiply the first coefficient 1 by the last term 6 to get %281%29%286%29=6.


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


To find these two numbers, we need to list all of the factors of 6 (the previous product).


Factors of 6:
1,2,3,6
-1,-2,-3,-6


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


These factors pair up and multiply to 6.
1*6
2*3
(-1)*(-6)
(-2)*(-3)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 7:


First NumberSecond NumberSum
161+6=7
232+3=5
-1-6-1+(-6)=-7
-2-3-2+(-3)=-5



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


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


Now replace the middle term 7x with x%2B6x. Remember, 1 and 6 add to 7. So this shows us that x%2B6x=7x.


x%5E2%2Bhighlight%28x%2B6x%29%2B6 Replace the second term 7x with x%2B6x.


%28x%5E2%2Bx%29%2B%286x%2B6%29 Group the terms into two pairs.


x%28x%2B1%29%2B%286x%2B6%29 Factor out the GCF x from the first group.


x%28x%2B1%29%2B6%28x%2B1%29 Factor out 6 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%2B6%29%28x%2B1%29 Combine like terms. Or factor out the common term x%2B1





So our expression goes from 7%28x%5E2%2B7x%2B6%29 and factors further to 7%28x%2B6%29%28x%2B1%29


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Answer:

So 7x%5E2%2B49x%2B42 factors to 7%28x%2B6%29%28x%2B1%29