SOLUTION: How to factor completely b^4 + 2a^2 b^2 + a^4 I have no idea how to factor this, there is no GCF, I don't see any groupings or any other type of factoring. Thank You

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: How to factor completely b^4 + 2a^2 b^2 + a^4 I have no idea how to factor this, there is no GCF, I don't see any groupings or any other type of factoring. Thank You       Log On


   

Question 626189: How to factor completely
b^4 + 2a^2 b^2 + a^4
I have no idea how to factor this, there is no GCF, I don't see any groupings or any other type of factoring.
Thank You



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

Looking at the expression b%5E4%2B2a%5E2b%5E2%2Ba%5E4, we can see that the first coefficient is 1, the second coefficient is 2, and the last coefficient is 1.


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


Now the question is: what two whole numbers multiply to 1 (the previous product) and add to the second coefficient 2?


To find these two numbers, we need to list all of the factors of 1 (the previous product).


Factors of 1:
1
-1


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


These factors pair up and multiply to 1.
1*1 = 1
(-1)*(-1) = 1

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


First NumberSecond NumberSum
111+1=2
-1-1-1+(-1)=-2



From the table, we can see that the two numbers 1 and 1 add to 2 (the middle coefficient).


So the two numbers 1 and 1 both multiply to 1 and add to 2


Now replace the middle term 2a%5E2b%5E2 with a%5E2b%5E2%2Ba%5E2b%5E2. Remember, 1 and 1 add to 2. So this shows us that a%5E2b%5E2%2Ba%5E2b%5E2=2a%5E2b%5E2.


b%5E4%2Bhighlight%28a%5E2b%5E2%2Ba%5E2b%5E2%29%2Ba%5E4 Replace the second term 2a%5E2b%5E2 with a%5E2b%5E2%2Ba%5E2b%5E2.


%28b%5E4%2Ba%5E2b%5E2%29%2B%28a%5E2b%5E2%2Ba%5E4%29 Group the terms into two pairs.


b%5E2%28b%5E2%2Ba%5E2%29%2B%28a%5E2b%5E2%2Ba%5E4%29 Factor out the GCF b%5E2 from the first group.


b%5E2%28b%5E2%2Ba%5E2%29%2Ba%5E2%28b%5E2%2Ba%5E2%29 Factor out a%5E2 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.


%28b%5E2%2Ba%5E2%29%28b%5E2%2Ba%5E2%29 Combine like terms. Or factor out the common term b%5E2%2Ba%5E2


%28b%5E2%2Ba%5E2%29%5E2 Condense the terms.


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Answer:


So b%5E4%2B2a%5E2b%5E2%2Ba%5E4 factors to %28b%5E2%2Ba%5E2%29%5E2.


In other words, b%5E4%2B2a%5E2b%5E2%2Ba%5E4=%28b%5E2%2Ba%5E2%29%5E2.


Note: you can check the answer by expanding %28b%5E2%2Ba%5E2%29%5E2 to get b%5E4%2B2a%5E2b%5E2%2Ba%5E4 or by graphing the original expression and the answer (the two graphs should be identical).

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