SOLUTION: Factor: z2 � 14z + 49 A) (z � 7)2 B) (z � 7)(z + 7) C) (z � 14)(z + 1) D) The expression is prime.

Question 140007: Factor: z2 � 14z + 49
A) (z � 7)2
B) (z � 7)(z + 7)
C) (z � 14)(z + 1)
D) The expression is prime.

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 1 and 49 respectively.

Now multiply the first coefficient 1 and the last coefficient 49 to get 49. Now what two numbers multiply to 49 and add to the middle coefficient -14? Let's list all of the factors of 49:

Factors of 49:
1,7

-1,-7 ...List the negative factors as well. This will allow us to find all possible combinations

These factors pair up and multiply to 49
1*49
7*7
(-1)*(-49)
(-7)*(-7)

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

Now which of these pairs add to -14? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -14

First Number	Second Number	Sum
1	49	1+49=50
7	7	7+7=14
-1	-49	-1+(-49)=-50
-7	-7	-7+(-7)=-14

From this list we can see that -7 and -7 add up to -14 and multiply to 49

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 )

note: is equivalent to since the term occurs twice. So also factors to

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

So factors to

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