Question 253891
{{{(x + 2y)^2-3(x+2y)+2}}} Start with the given expression.



Let {{{z=x+2y}}}



{{{z^2-3z+2}}} Replace {{{x+2y}}} with 'z'



Looking at the expression {{{z^2-3z+2}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-3}}}, and the last term is {{{2}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{2}}} to get {{{(1)(2)=2}}}.



Now the question is: what two whole numbers multiply to {{{2}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-3}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{2}}} (the previous product).



Factors of {{{2}}}:

1,2

-1,-2



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



These factors pair up and multiply to {{{2}}}.

1*2 = 2
(-1)*(-2) = 2


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-3}}}:



<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>2</font></td><td  align="center"><font color=black>1+2=3</font></td></tr><tr><td  align="center"><font color=red>-1</font></td><td  align="center"><font color=red>-2</font></td><td  align="center"><font color=red>-1+(-2)=-3</font></td></tr></table>



From the table, we can see that the two numbers {{{-1}}} and {{{-2}}} add to {{{-3}}} (the middle coefficient).



So the two numbers {{{-1}}} and {{{-2}}} both multiply to {{{2}}} <font size=4><b>and</b></font> add to {{{-3}}}



Now replace the middle term {{{-3z}}} with {{{-z-2z}}}. Remember, {{{-1}}} and {{{-2}}} add to {{{-3}}}. So this shows us that {{{-z-2z=-3z}}}.



{{{z^2+highlight(-z-2z)+2}}} Replace the second term {{{-3z}}} with {{{-z-2z}}}.



{{{(z^2-z)+(-2z+2)}}} Group the terms into two pairs.



{{{z(z-1)+(-2z+2)}}} Factor out the GCF {{{z}}} from the first group.



{{{z(z-1)-2(z-1)}}} Factor out {{{2}}} 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.



{{{(z-2)(z-1)}}} Combine like terms. Or factor out the common term {{{z-1}}}



{{{(x+2y-2)(x+2y-1)}}} Plug in {{{z=x+2y}}}


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Answer:



So {{{(x + 2y)^2-3(x+2y)+2}}} factors to {{{(x+2y-2)(x+2y-1)}}}.



In other words, {{{(x + 2y)^2-3(x+2y)+2=(x+2y-2)(x+2y-1)}}}.