SOLUTION: Factor completely: 7x^2 + 58x + 16

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Question 145894: Factor completely:
7x^2 + 58x + 16

Found 2 solutions by scott8148, jim_thompson5910:
Answer by scott8148(6628)   (Show Source): You can put this solution on YOUR website!
(7x+2)(x+8)
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

Looking at we can see that the first term is and the last term is where the coefficients are 7 and 16 respectively.

Now multiply the first coefficient 7 and the last coefficient 16 to get 112. Now what two numbers multiply to 112 and add to the middle coefficient 58? Let's list all of the factors of 112:



Factors of 112:
1,2,4,7,8,14,16,28,56,112

-1,-2,-4,-7,-8,-14,-16,-28,-56,-112 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to 112
1*112
2*56
4*28
7*16
8*14
(-1)*(-112)
(-2)*(-56)
(-4)*(-28)
(-7)*(-16)
(-8)*(-14)

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


Now which of these pairs add to 58? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 58

First NumberSecond NumberSum
11121+112=113
2562+56=58
4284+28=32
7167+16=23
8148+14=22
-1-112-1+(-112)=-113
-2-56-2+(-56)=-58
-4-28-4+(-28)=-32
-7-16-7+(-16)=-23
-8-14-8+(-14)=-22



From this list we can see that 2 and 56 add up to 58 and multiply to 112


Now looking at the expression , replace with (notice adds up to . So it is equivalent to )




Now let's factor by grouping:


Group like terms


Factor out the GCF of out of the first group. Factor out the GCF of out of the second group


Since we have a common term of , we can combine like terms

So factors to


So this also means that factors to (since is equivalent to )


-------------------------------
Answer:

So factors to

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