Question 131050


Looking at {{{1z^2+5z-84}}} we can see that the first term is {{{1z^2}}} and the last term is {{{-84}}} where the coefficients are 1 and -84 respectively.


Now multiply the first coefficient 1 and the last coefficient -84 to get -84. Now what two numbers multiply to -84 and add to the  middle coefficient 5? Let's list all of the factors of -84:




Factors of -84:

1,2,3,4,6,7,12,14,21,28,42,84


-1,-2,-3,-4,-6,-7,-12,-14,-21,-28,-42,-84 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -84

(1)*(-84)

(2)*(-42)

(3)*(-28)

(4)*(-21)

(6)*(-14)

(7)*(-12)

(-1)*(84)

(-2)*(42)

(-3)*(28)

(-4)*(21)

(-6)*(14)

(-7)*(12)


note: remember, the product of a negative and a positive number is a negative number



Now which of these pairs add to 5? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 5


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-84</td><td>1+(-84)=-83</td></tr><tr><td align="center">2</td><td align="center">-42</td><td>2+(-42)=-40</td></tr><tr><td align="center">3</td><td align="center">-28</td><td>3+(-28)=-25</td></tr><tr><td align="center">4</td><td align="center">-21</td><td>4+(-21)=-17</td></tr><tr><td align="center">6</td><td align="center">-14</td><td>6+(-14)=-8</td></tr><tr><td align="center">7</td><td align="center">-12</td><td>7+(-12)=-5</td></tr><tr><td align="center">-1</td><td align="center">84</td><td>-1+84=83</td></tr><tr><td align="center">-2</td><td align="center">42</td><td>-2+42=40</td></tr><tr><td align="center">-3</td><td align="center">28</td><td>-3+28=25</td></tr><tr><td align="center">-4</td><td align="center">21</td><td>-4+21=17</td></tr><tr><td align="center">-6</td><td align="center">14</td><td>-6+14=8</td></tr><tr><td align="center">-7</td><td align="center">12</td><td>-7+12=5</td></tr></table>



From this list we can see that -7 and 12 add up to 5 and multiply to -84



Now looking at the expression {{{1z^2+5z-84}}}, replace {{{5z}}} with {{{-7z+12z}}} (notice {{{-7z+12z}}} adds up to {{{5z}}}. So it is equivalent to {{{5z}}})


{{{1z^2+highlight(-7z+12z)+-84}}}



Now let's factor {{{1z^2-7z+12z-84}}} by grouping:



{{{(1z^2-7z)+(12z-84)}}} Group like terms



{{{z(z-7)+12(z-7)}}} Factor out the GCF of {{{z}}} out of the first group. Factor out the GCF of {{{12}}} out of the second group



{{{(z+12)(z-7)}}} Since we have a common term of {{{z-7}}}, we can combine like terms


So {{{1z^2-7z+12z-84}}} factors to {{{(z+12)(z-7)}}}



So this also means that {{{1z^2+5z-84}}} factors to {{{(z+12)(z-7)}}} (since {{{1z^2+5z-84}}} is equivalent to {{{1z^2-7z+12z-84}}})




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     Answer:

So {{{z^2+5z-84}}} factors to {{{(z+12)(z-7)}}} which means that one of the factors are either {{{z+12}}} or {{{z-7}}}