SOLUTION: Factor the trinomial completely 7a^2+50ab+7b^2

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Question 191472: Factor the trinomial completely
7a^2+50ab+7b^2

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

Looking at 7a%5E2%2B50ab%2B7b%5E2 we can see that the first term is 7a%5E2 and the last term is 7b%5E2 where the coefficients are 7 and 7 respectively.

Now multiply the first coefficient 7 and the last coefficient 7 to get 49. Now what two numbers multiply to 49 and add to the middle coefficient 50? Let's list all of the factors of 49:



Factors of 49:
1,7

-1,-7 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to 49
1*49
7*7
(-1)*(-49)
(-7)*(-7)

note: remember two negative numbers multiplied together make a positive number


Now which of these pairs add to 50? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 50

First NumberSecond NumberSum
1491+49=50
777+7=14
-1-49-1+(-49)=-50
-7-7-7+(-7)=-14



From this list we can see that 1 and 49 add up to 50 and multiply to 49


Now looking at the expression 7a%5E2%2B50ab%2B7b%5E2, replace 50ab with ab%2B49ab (notice ab%2B49ab adds up to 50ab. So it is equivalent to 50ab)

7a%5E2%2Bhighlight%28ab%2B49ab%29%2B7b%5E2


Now let's factor 7a%5E2%2Bab%2B49ab%2B7b%5E2 by grouping:


%287a%5E2%2Bab%29%2B%2849ab%2B7b%5E2%29 Group like terms


a%287a%2Bb%29%2B7b%287a%2Bb%29 Factor out the GCF of a out of the first group. Factor out the GCF of 7b out of the second group


%28a%2B7b%29%287a%2Bb%29 Since we have a common term of 7a%2Bb, we can combine like terms

So 7a%5E2%2B1ab%2B49ab%2B7b%5E2 factors to %28a%2B7b%29%287a%2Bb%29


So this also means that 7a%5E2%2B50ab%2B7b%5E2 factors to %28a%2B7b%29%287a%2Bb%29 (since 7a%5E2%2B50ab%2B7b%5E2 is equivalent to 7a%5E2%2B1ab%2B49ab%2B7b%5E2)



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Answer:
So 7a%5E2%2B50ab%2B7b%5E2 factors to %28a%2B7b%29%287a%2Bb%29