SOLUTION: factor completely x^2+22x+57

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Question 191799: factor completely

x^2+22x+57

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

Looking at the expression x%5E2%2B22x%2B57, we can see that the first coefficient is 1, the second coefficient is 22, and the last term is 57.


Now multiply the first coefficient 1 by the last term 57 to get %281%29%2857%29=57.


Now the question is: what two whole numbers multiply to 57 (the previous product) and add to the second coefficient 22?


To find these two numbers, we need to list all of the factors of 57 (the previous product).


Factors of 57:
1,3,19,57
-1,-3,-19,-57


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


These factors pair up and multiply to 57.
1*57
3*19
(-1)*(-57)
(-3)*(-19)

Now let's add up each pair of factors to see if one pair adds to the middle coefficient 22:


First NumberSecond NumberSum
1571+57=58
3193+19=22
-1-57-1+(-57)=-58
-3-19-3+(-19)=-22



From the table, we can see that the two numbers 3 and 19 add to 22 (the middle coefficient).


So the two numbers 3 and 19 both multiply to 57 and add to 22


Now replace the middle term 22x with 3x%2B19x. Remember, 3 and 19 add to 22. So this shows us that 3x%2B19x=22x.


x%5E2%2Bhighlight%283x%2B19x%29%2B57 Replace the second term 22x with 3x%2B19x.


%28x%5E2%2B3x%29%2B%2819x%2B57%29 Group the terms into two pairs.


x%28x%2B3%29%2B%2819x%2B57%29 Factor out the GCF x from the first group.


x%28x%2B3%29%2B19%28x%2B3%29 Factor out 19 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%2B19%29%28x%2B3%29 Combine like terms. Or factor out the common term x%2B3

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Answer:


So x%5E2%2B22x%2B57 factors to %28x%2B19%29%28x%2B3%29.


Note: you can check the answer by FOILing %28x%2B19%29%28x%2B3%29 to get x%5E2%2B22x%2B57 or by graphing the original expression and the answer (the two graphs should be identical).