SOLUTION: factor completely 2b^2-11b+5

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Question 421297: factor completely
2b^2-11b+5
Answer by MathLover1(20849) About Me  (Show Source):
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Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 2x%5E2-11x%2B5, we can see that the first coefficient is 2, the second coefficient is -11, and the last term is 5.



Now multiply the first coefficient 2 by the last term 5 to get %282%29%285%29=10.



Now the question is: what two whole numbers multiply to 10 (the previous product) and add to the second coefficient -11?



To find these two numbers, we need to list all of the factors of 10 (the previous product).



Factors of 10:

1,2,5,10

-1,-2,-5,-10



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



These factors pair up and multiply to 10.

1*10 = 10
2*5 = 10
(-1)*(-10) = 10
(-2)*(-5) = 10


Now let's add up each pair of factors to see if one pair adds to the middle coefficient -11:



First NumberSecond NumberSum
1101+10=11
252+5=7
-1-10-1+(-10)=-11
-2-5-2+(-5)=-7




From the table, we can see that the two numbers -1 and -10 add to -11 (the middle coefficient).



So the two numbers -1 and -10 both multiply to 10 and add to -11



Now replace the middle term -11x with -x-10x. Remember, -1 and -10 add to -11. So this shows us that -x-10x=-11x.



2x%5E2%2Bhighlight%28-x-10x%29%2B5 Replace the second term -11x with -x-10x.



%282x%5E2-x%29%2B%28-10x%2B5%29 Group the terms into two pairs.



x%282x-1%29%2B%28-10x%2B5%29 Factor out the GCF x from the first group.



x%282x-1%29-5%282x-1%29 Factor out 5 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-5%29%282x-1%29 Combine like terms. Or factor out the common term 2x-1



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Answer:



So 2%2Ax%5E2-11%2Ax%2B5 factors to %28x-5%29%282x-1%29.



In other words, 2%2Ax%5E2-11%2Ax%2B5=%28x-5%29%282x-1%29.



Note: you can check the answer by expanding %28x-5%29%282x-1%29 to get 2%2Ax%5E2-11%2Ax%2B5 or by graphing the original expression and the answer (the two graphs should be identical).