SOLUTION: factor completely 112b^2+392bg+343g^2

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Question 515414: factor completely 112b^2+392bg+343g^2
Answer by drcole(72) About Me  (Show Source):
You can put this solution on YOUR website!
To factor 112+b%5E2+%2B+392+b+g+%2B+343+g%5E2+, we first notice that all of the coefficients are divisible by 7, so we factor 7 out of each term:
+7%2816+b%5E2+%2B+56++b+g+%2B+49+g%5E2%29+
There are no other common factors of the coefficients. Now we need to find two numbers whose product is (16)(49) = 784 and whose sum is 56. Let's write out some possible ways to factor 784 and look at the sum of the factors:
(1)(784) sum = 785
(2)(392) sum = 394
(4)(196) sum = 200
(7)(112) sum = 119
(8)(98) sum = 106
(14)(56) sum = 70
(16)(49) sum = 65
(28)(28) sum = 56 <--- this works!
So we've found our combination. We can rewrite the quadratic expression using these two factors:
7%2816+b%5E2+%2B+28+b+g+%2B+28+b+g+%2B+343+g%5E2%29
Now we can factor the quadratic by grouping. The greatest common factor of 16+b%5E2 and 28+b+g is 4b, so we factor that out of the first two terms:
7%284b%28+4b+%2B+7g%29+%2B+28+b+g+%2B+343+g%5E2%29
The greatest common factor of 28+b+g and 343+g%5E2 is 7g, so we factor this out of the last two terms:
7%284b%284b+%2B+7g%29+%2B+7g%284b+%2B+7g%29%29
Now we factor 4b+%2B+7g out of both terms:
7%284b+%2B+7g%29%284b+%2B+7g%29+=+7%284b+%2B+7g%29%5E2
This completes the factorization.