Question 118316


Looking at {{{2m^2+11mn+9n^2}}} we can see that the first term is {{{2m^2}}} and the last term is {{{9n^2}}} where the coefficients are 2 and 9 respectively.


Now multiply the first coefficient 2 and the last coefficient 9 to get 18. Now what two numbers multiply to 18 and add to the  middle coefficient 11? Let's list all of the factors of 18:




Factors of 18:

1,2,3,6,9,18


-1,-2,-3,-6,-9,-18 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 18

1*18

2*9

3*6

(-1)*(-18)

(-2)*(-9)

(-3)*(-6)


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



Now which of these pairs add to 11? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 11


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">18</td><td>1+18=19</td></tr><tr><td align="center">2</td><td align="center">9</td><td>2+9=11</td></tr><tr><td align="center">3</td><td align="center">6</td><td>3+6=9</td></tr><tr><td align="center">-1</td><td align="center">-18</td><td>-1+(-18)=-19</td></tr><tr><td align="center">-2</td><td align="center">-9</td><td>-2+(-9)=-11</td></tr><tr><td align="center">-3</td><td align="center">-6</td><td>-3+(-6)=-9</td></tr></table>



From this list we can see that 2 and 9 add up to 11 and multiply to 18



Now looking at the expression {{{2m^2+11mn+9n^2}}}, replace {{{11mn}}} with {{{2mn+9mn}}} (notice {{{2mn+9mn}}} adds up to {{{11mn}}}. So it is equivalent to {{{11mn}}})


{{{2m^2+highlight(2mn+9mn)+9n^2}}}



Now let's factor {{{2m^2+2mn+9mn+9n^2}}} by grouping:



{{{(2m^2+2mn)+(9mn+9n^2)}}} Group like terms



{{{2m(m+n)+9n(m+n)}}} Factor out the GCF of {{{2m}}} out of the first group. Factor out the GCF of {{{9n}}} out of the second group



{{{(2m+9n)(m+n)}}} Since we have a common term of {{{m+n}}}, we can combine like terms


So {{{2m^2+2mn+9mn+9n^2}}} factors to {{{(2m+9n)(m+n)}}}



So this also means that {{{2m^2+11mn+9n^2}}} factors to {{{(2m+9n)(m+n)}}} (since {{{2m^2+11mn+9n^2}}} is equivalent to {{{2m^2+2mn+9mn+9n^2}}})


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Answer:


So {{{2m^2+11mn+9n^2}}} factors to {{{(2m+9n)(m+n)}}}