Question 332267

Looking at the expression {{{x^2+12xy+36y^2}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{12}}}, and the last coefficient is {{{36}}}.



Now multiply the first coefficient {{{1}}} by the last coefficient {{{36}}} to get {{{(1)(36)=36}}}.



Now the question is: what two whole numbers multiply to {{{36}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{12}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{36}}} (the previous product).



Factors of {{{36}}}:

1,2,3,4,6,9,12,18,36

-1,-2,-3,-4,-6,-9,-12,-18,-36



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



These factors pair up and multiply to {{{36}}}.

1*36 = 36
2*18 = 36
3*12 = 36
4*9 = 36
6*6 = 36
(-1)*(-36) = 36
(-2)*(-18) = 36
(-3)*(-12) = 36
(-4)*(-9) = 36
(-6)*(-6) = 36


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{12}}}:



<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>36</font></td><td  align="center"><font color=black>1+36=37</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>18</font></td><td  align="center"><font color=black>2+18=20</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>12</font></td><td  align="center"><font color=black>3+12=15</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>9</font></td><td  align="center"><font color=black>4+9=13</font></td></tr><tr><td  align="center"><font color=red>6</font></td><td  align="center"><font color=red>6</font></td><td  align="center"><font color=red>6+6=12</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-36</font></td><td  align="center"><font color=black>-1+(-36)=-37</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-18</font></td><td  align="center"><font color=black>-2+(-18)=-20</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-12</font></td><td  align="center"><font color=black>-3+(-12)=-15</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>-9</font></td><td  align="center"><font color=black>-4+(-9)=-13</font></td></tr><tr><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>-6+(-6)=-12</font></td></tr></table>



From the table, we can see that the two numbers {{{6}}} and {{{6}}} add to {{{12}}} (the middle coefficient).



So the two numbers {{{6}}} and {{{6}}} both multiply to {{{36}}} <font size=4><b>and</b></font> add to {{{12}}}



Now replace the middle term {{{12xy}}} with {{{6xy+6xy}}}. Remember, {{{6}}} and {{{6}}} add to {{{12}}}. So this shows us that {{{6xy+6xy=12xy}}}.



{{{x^2+highlight(6xy+6xy)+36y^2}}} Replace the second term {{{12xy}}} with {{{6xy+6xy}}}.



{{{(x^2+6xy)+(6xy+36y^2)}}} Group the terms into two pairs.



{{{x(x+6y)+(6xy+36y^2)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x+6y)+6y(x+6y)}}} Factor out {{{6y}}} 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.



{{{(x+6y)(x+6y)}}} Combine like terms. Or factor out the common term {{{x+6y}}}



{{{(x+6y)^2}}} Condense the terms.



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Answer:



So {{{x^2+12xy+36y^2}}} factors to {{{(x+6y)^2}}}.



In other words, {{{x^2+12xy+36y^2=(x+6y)^2}}}.