Question 190285


Looking at {{{x^2+8xy+12y^2}}} we can see that the first term is {{{x^2}}} and the last term is {{{12y^2}}} where the coefficients are 1 and 12 respectively.


Now multiply the first coefficient 1 and the last coefficient 12 to get 12. Now what two numbers multiply to 12 and add to the  middle coefficient 8? Let's list all of the factors of 12:




Factors of 12:

1,2,3,4,6,12


-1,-2,-3,-4,-6,-12 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to 12

1*12

2*6

3*4

(-1)*(-12)

(-2)*(-6)

(-3)*(-4)


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



Now which of these pairs add to 8? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 8


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">12</td><td>1+12=13</td></tr><tr><td align="center">2</td><td align="center">6</td><td>2+6=8</td></tr><tr><td align="center">3</td><td align="center">4</td><td>3+4=7</td></tr><tr><td align="center">-1</td><td align="center">-12</td><td>-1+(-12)=-13</td></tr><tr><td align="center">-2</td><td align="center">-6</td><td>-2+(-6)=-8</td></tr><tr><td align="center">-3</td><td align="center">-4</td><td>-3+(-4)=-7</td></tr></table>



From this list we can see that 2 and 6 add up to 8 and multiply to 12



Now looking at the expression {{{x^2+8xy+12y^2}}}, replace {{{8xy}}} with {{{2xy+6xy}}} (notice {{{2xy+6xy}}} adds up to {{{8xy}}}. So it is equivalent to {{{8xy}}})


{{{x^2+highlight(2xy+6xy)+12y^2}}}



Now let's factor {{{x^2+2xy+6xy+12y^2}}} by grouping:



{{{(x^2+2xy)+(6xy+12y^2)}}} Group like terms



{{{x(x+2y)+6y(x+2y)}}} Factor out the GCF of {{{x}}} out of the first group. Factor out the GCF of {{{6y}}} out of the second group



{{{(x+6y)(x+2y)}}} Since we have a common term of {{{x+2y}}}, we can combine like terms


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



So this also means that {{{x^2+8xy+12y^2}}} factors to {{{(x+6y)(x+2y)}}} (since {{{x^2+8xy+12y^2}}} is equivalent to {{{x^2+2xy+6xy+12y^2}}})




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




     Answer:

So {{{x^2+8xy+12y^2}}} factors to {{{(x+6y)(x+2y)}}}