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

Algebra ->  Test -> SOLUTION: Factor: z2 – 14z + 49 A) (z – 7)2 B) (z – 7)(z + 7) C) (z – 14)(z + 1) D) The expression is prime.       Log On


   



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

Looking at 1z%5E2-14z%2B49 we can see that the first term is 1z%5E2 and the last term is 49 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 NumberSecond NumberSum
1491+49=50
777+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 1z%5E2-14z%2B49, replace -14z with -7z%2B-7z (notice -7z%2B-7z adds up to -14z. So it is equivalent to -14z)

1z%5E2%2Bhighlight%28-7z%2B-7z%29%2B49


Now let's factor 1z%5E2-7z-7z%2B49 by grouping:


%281z%5E2-7z%29%2B%28-7z%2B49%29 Group like terms


z%28z-7%29-7%28z-7%29 Factor out the GCF of z out of the first group. Factor out the GCF of -7 out of the second group


%28z-7%29%28z-7%29 Since we have a common term of z-7, we can combine like terms

So 1z%5E2-7z-7z%2B49 factors to %28z-7%29%28z-7%29


So this also means that 1z%5E2-14z%2B49 factors to %28z-7%29%28z-7%29 (since 1z%5E2-14z%2B49 is equivalent to 1z%5E2-7z-7z%2B49)


note: %28z-7%29%28z-7%29 is equivalent to %28z-7%29%5E2 since the term z-7 occurs twice. So 1z%5E2-14z%2B49 also factors to %28z-7%29%5E2


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Answer:

So 1z%5E2-14z%2B49 factors to %28z-7%29%5E2