Question 140007


Looking at {{{1z^2-14z+49}}} we can see that the first term is {{{1z^2}}} 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


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">49</td><td>1+49=50</td></tr><tr><td align="center">7</td><td align="center">7</td><td>7+7=14</td></tr><tr><td align="center">-1</td><td align="center">-49</td><td>-1+(-49)=-50</td></tr><tr><td align="center">-7</td><td align="center">-7</td><td>-7+(-7)=-14</td></tr></table>



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



Now looking at the expression {{{1z^2-14z+49}}}, replace {{{-14z}}} with {{{-7z+-7z}}} (notice {{{-7z+-7z}}} adds up to {{{-14z}}}. So it is equivalent to {{{-14z}}})


{{{1z^2+highlight(-7z+-7z)+49}}}



Now let's factor {{{1z^2-7z-7z+49}}} by grouping:



{{{(1z^2-7z)+(-7z+49)}}} Group like terms



{{{z(z-7)-7(z-7)}}} Factor out the GCF of {{{z}}} out of the first group. Factor out the GCF of {{{-7}}} out of the second group



{{{(z-7)(z-7)}}} Since we have a common term of {{{z-7}}}, we can combine like terms


So {{{1z^2-7z-7z+49}}} factors to {{{(z-7)(z-7)}}}



So this also means that {{{1z^2-14z+49}}} factors to {{{(z-7)(z-7)}}} (since {{{1z^2-14z+49}}} is equivalent to {{{1z^2-7z-7z+49}}})



note:  {{{(z-7)(z-7)}}} is equivalent to  {{{(z-7)^2}}} since the term {{{z-7}}} occurs twice. So {{{1z^2-14z+49}}} also factors to {{{(z-7)^2}}}



-------------------------------

Answer:


So {{{1z^2-14z+49}}} factors to {{{(z-7)^2}}}