Question 187009
{{{12z^2+z=6}}} Start with the given equation.



{{{12z^2+z-6=0}}} Subtract 6 from both sides.



Let's factor the left side:



Looking at the expression {{{12z^2+z-6}}}, we can see that the first coefficient is {{{12}}}, the second coefficient is {{{1}}}, and the last term is {{{-6}}}.



Now multiply the first coefficient {{{12}}} by the last term {{{-6}}} to get {{{(12)(-6)=-72}}}.



Now the question is: what two whole numbers multiply to {{{-72}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{1}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-72}}} (the previous product).



Factors of {{{-72}}}:

1,2,3,4,6,8,9,12,18,24,36,72

-1,-2,-3,-4,-6,-8,-9,-12,-18,-24,-36,-72



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{-72}}}.

1*(-72)
2*(-36)
3*(-24)
4*(-18)
6*(-12)
8*(-9)
(-1)*(72)
(-2)*(36)
(-3)*(24)
(-4)*(18)
(-6)*(12)
(-8)*(9)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{1}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>-72</font></td><td  align="center"><font color=black>1+(-72)=-71</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>-36</font></td><td  align="center"><font color=black>2+(-36)=-34</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>-24</font></td><td  align="center"><font color=black>3+(-24)=-21</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>-18</font></td><td  align="center"><font color=black>4+(-18)=-14</font></td></tr><tr><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>-12</font></td><td  align="center"><font color=black>6+(-12)=-6</font></td></tr><tr><td  align="center"><font color=black>8</font></td><td  align="center"><font color=black>-9</font></td><td  align="center"><font color=black>8+(-9)=-1</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>72</font></td><td  align="center"><font color=black>-1+72=71</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>36</font></td><td  align="center"><font color=black>-2+36=34</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>24</font></td><td  align="center"><font color=black>-3+24=21</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>18</font></td><td  align="center"><font color=black>-4+18=14</font></td></tr><tr><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>12</font></td><td  align="center"><font color=black>-6+12=6</font></td></tr><tr><td  align="center"><font color=red>-8</font></td><td  align="center"><font color=red>9</font></td><td  align="center"><font color=red>-8+9=1</font></td></tr></table>



From the table, we can see that the two numbers {{{-8}}} and {{{9}}} add to {{{1}}} (the middle coefficient).



So the two numbers {{{-8}}} and {{{9}}} both multiply to {{{-72}}} <font size=4><b>and</b></font> add to {{{1}}}



Now replace the middle term {{{1z}}} with {{{-8z+9z}}}. Remember, {{{-8}}} and {{{9}}} add to {{{1}}}. So this shows us that {{{-8z+9z=1z}}}.



{{{12z^2+highlight(-8z+9z)-6}}} Replace the second term {{{1z}}} with {{{-8z+9z}}}.



{{{(12z^2-8z)+(9z-6)}}} Group the terms into two pairs.



{{{4z(3z-2)+(9z-6)}}} Factor out the GCF {{{4z}}} from the first group.



{{{4z(3z-2)+3(3z-2)}}} Factor out {{{3}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(4z+3)(3z-2)}}} Combine like terms. Or factor out the common term {{{3z-2}}}



So {{{12z^2+z-6}}} factors to {{{(4z+3)(3z-2)}}}.



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



So {{{12z^2+z-6=0}}} becomes {{{(4z+3)(3z-2)=0}}}



Now set each factor equal to zero:

{{{4z+3=0}}} or  {{{3z-2=0}}} 


{{{z=-3/4}}} or  {{{z=2/3}}}    Now solve for z in each case



So our answers are 


 {{{z=-3/4}}} or  {{{z=2/3}}}