Question 333713


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



Now multiply the first coefficient {{{12}}} by the last coefficient {{{-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) = -72
2*(-36) = -72
3*(-24) = -72
4*(-18) = -72
6*(-12) = -72
8*(-9) = -72
(-1)*(72) = -72
(-2)*(36) = -72
(-3)*(24) = -72
(-4)*(18) = -72
(-6)*(12) = -72
(-8)*(9) = -72


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 {{{1xy}}} with {{{-8xy+9xy}}}. Remember, {{{-8}}} and {{{9}}} add to {{{1}}}. So this shows us that {{{-8xy+9xy=1xy}}}.



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



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



{{{4x(3x-2y)+(9xy-6y^2)}}} Factor out the GCF {{{4x}}} from the first group.



{{{4x(3x-2y)+3y(3x-2y)}}} Factor out {{{3y}}} 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.



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



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Answer:



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



In other words, {{{12x^2+xy-6y^2=(4x+3y)(3x-2y)}}}.



Note: you can check the answer by expanding {{{(4x+3y)(3x-2y)}}} to get {{{12x^2+xy-6y^2}}} or by graphing the original expression and the answer (the two graphs should be identical).