Question 905066

{{{108x^2+63xy+9y^2}}} Start with the given expression.



{{{9(12x^2+7xy+y^2)}}} Factor out the GCF {{{9}}}.



Now let's try to factor the inner expression {{{12x^2+7xy+y^2}}}



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Looking at the expression {{{12x^2+7xy+y^2}}}, we can see that the first coefficient is {{{12}}}, the second coefficient is {{{7}}}, and the last coefficient is {{{1}}}.



Now multiply the first coefficient {{{12}}} by the last coefficient {{{1}}} to get {{{(12)(1)=12}}}.



Now the question is: what two whole numbers multiply to {{{12}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{7}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{12}}} (the previous product).



Factors of {{{12}}}:

1,2,3,4,6,12

-1,-2,-3,-4,-6,-12



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



These factors pair up and multiply to {{{12}}}.

1*12 = 12
2*6 = 12
3*4 = 12
(-1)*(-12) = 12
(-2)*(-6) = 12
(-3)*(-4) = 12


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{7}}}:



<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>12</font></td><td  align="center"><font color=black>1+12=13</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>2+6=8</font></td></tr><tr><td  align="center"><font color=red>3</font></td><td  align="center"><font color=red>4</font></td><td  align="center"><font color=red>3+4=7</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-12</font></td><td  align="center"><font color=black>-1+(-12)=-13</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>-2+(-6)=-8</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>-3+(-4)=-7</font></td></tr></table>



From the table, we can see that the two numbers {{{3}}} and {{{4}}} add to {{{7}}} (the middle coefficient).



So the two numbers {{{3}}} and {{{4}}} both multiply to {{{12}}} <font size=4><b>and</b></font> add to {{{7}}}



Now replace the middle term {{{7xy}}} with {{{3xy+4xy}}}. Remember, {{{3}}} and {{{4}}} add to {{{7}}}. So this shows us that {{{3xy+4xy=7xy}}}.



{{{12x^2+highlight(3xy+4xy)+y^2}}} Replace the second term {{{7xy}}} with {{{3xy+4xy}}}.



{{{(12x^2+3xy)+(4xy+y^2)}}} Group the terms into two pairs.



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



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



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



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So {{{9(12x^2+7xy+y^2)}}} then factors further to {{{9(3x+y)(4x+y)}}}



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Answer:



So {{{108x^2+63xy+9y^2}}} completely factors to {{{9(3x+y)(4x+y)}}}.



In other words, {{{108x^2+63xy+9y^2=9(3x+y)(4x+y)}}}.



Note: you can check the answer by expanding {{{9(3x+y)(4x+y)}}} to get {{{108x^2+63xy+9y^2}}}.