Question 299375


{{{2y^3+24y^2+72y}}} Start with the given expression.



{{{2y(y^2+12y+36)}}} Factor out the GCF {{{2y}}}.



Now let's try to factor the inner expression {{{y^2+12y+36}}}



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



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



Now multiply the first coefficient {{{1}}} by the last term {{{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 {{{12y}}} with {{{6y+6y}}}. Remember, {{{6}}} and {{{6}}} add to {{{12}}}. So this shows us that {{{6y+6y=12y}}}.



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



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



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



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



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



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



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



So {{{2y(y^2+12y+36)}}} then factors further to {{{2y(y+6)^2}}}



===============================================================



Answer:



So {{{2y^3+24y^2+72y}}} completely factors to {{{2y(y+6)^2}}}.



In other words, {{{2y^3+24y^2+72y=2y(y+6)^2}}}.



Note: you can check the answer by expanding {{{2y(y+6)^2}}} to get {{{2y^3+24y^2+72y}}} or by graphing the original expression and the answer (the two graphs should be identical).