Question 191474


Looking at {{{x^2+18xy+81y^2}}} we can see that the first term is {{{x^2}}} and the last term is {{{81y^2}}} where the coefficients are 1 and 81 respectively.


Now multiply the first coefficient 1 and the last coefficient 81 to get 81. Now what two numbers multiply to 81 and add to the  middle coefficient 18? Let's list all of the factors of 81:




Factors of 81:

1,3,9,27


-1,-3,-9,-27 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 81

1*81

3*27

9*9

(-1)*(-81)

(-3)*(-27)

(-9)*(-9)


note: remember two negative numbers multiplied together make a positive number



Now which of these pairs add to 18? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 18


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">81</td><td>1+81=82</td></tr><tr><td align="center">3</td><td align="center">27</td><td>3+27=30</td></tr><tr><td align="center">9</td><td align="center">9</td><td>9+9=18</td></tr><tr><td align="center">-1</td><td align="center">-81</td><td>-1+(-81)=-82</td></tr><tr><td align="center">-3</td><td align="center">-27</td><td>-3+(-27)=-30</td></tr><tr><td align="center">-9</td><td align="center">-9</td><td>-9+(-9)=-18</td></tr></table>



From this list we can see that 9 and 9 add up to 18 and multiply to 81



Now looking at the expression {{{x^2+18xy+81y^2}}}, replace {{{18xy}}} with {{{9xy+9xy}}} (notice {{{9xy+9xy}}} adds up to {{{18xy}}}. So it is equivalent to {{{18xy}}})


{{{x^2+highlight(9xy+9xy)+81y^2}}}



Now let's factor {{{x^2+9xy+9xy+81y^2}}} by grouping:



{{{(x^2+9xy)+(9xy+81y^2)}}} Group like terms



{{{x(x+9y)+9y(x+9y)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{9y}}} out of the second group



{{{(x+9y)(x+9y)}}} Since we have a common term of {{{x+9y}}}, we can combine like terms


So {{{x^2+9xy+9xy+81y^2}}} factors to {{{(x+9y)(x+9y)}}}



So this also means that {{{x^2+18xy+81y^2}}} factors to {{{(x+9y)(x+9y)}}} (since {{{x^2+18xy+81y^2}}} is equivalent to {{{x^2+9xy+9xy+81y^2}}})



note:  {{{(x+9y)(x+9y)}}} is equivalent to  {{{(x+9y)^2}}} since the term {{{x+9y}}} occurs twice. So {{{x^2+18xy+81y^2}}} also factors to {{{(x+9y)^2}}}




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     Answer:

So {{{x^2+18xy+81y^2}}} factors to {{{(x+9y)^2}}}