Question 151358
Let {{{z=3x+y}}} 


So the expression goes from {{{5(3x+y)^2+6(3x+y)-8}}} to {{{5z^2+6z-8}}}




Looking at the expression {{{5z^2+6z-8}}}, we can see that the first coefficient is {{{5}}}, the second coefficient is {{{6}}}, and the last term is {{{-8}}}.



Now multiply the first coefficient {{{5}}} by the last term {{{-8}}} to get {{{(5)(-8)=-40}}}.



Now the question is: what two whole numbers multiply to {{{-40}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{6}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-40}}} (the previous product).



Factors of {{{-40}}}:

1,2,4,5,8,10,20,40

-1,-2,-4,-5,-8,-10,-20,-40



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



These factors pair up and multiply to {{{-40}}}.

1*(-40)
2*(-20)
4*(-10)
5*(-8)
(-1)*(40)
(-2)*(20)
(-4)*(10)
(-5)*(8)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{6}}}:



<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>-40</font></td><td  align="center"><font color=black>1+(-40)=-39</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>-20</font></td><td  align="center"><font color=black>2+(-20)=-18</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>4+(-10)=-6</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>-8</font></td><td  align="center"><font color=black>5+(-8)=-3</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>40</font></td><td  align="center"><font color=black>-1+40=39</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>20</font></td><td  align="center"><font color=black>-2+20=18</font></td></tr><tr><td  align="center"><font color=red>-4</font></td><td  align="center"><font color=red>10</font></td><td  align="center"><font color=red>-4+10=6</font></td></tr><tr><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>8</font></td><td  align="center"><font color=black>-5+8=3</font></td></tr></table>



From the table, we can see that the two numbers {{{-4}}} and {{{10}}} add to {{{6}}} (the middle coefficient).



So the two numbers {{{-4}}} and {{{10}}} both multiply to {{{-40}}} <font size=4><b>and</b></font> add to {{{6}}}



Now replace the middle term {{{6z}}} with {{{-4z+10z}}}. Remember, {{{-4}}} and {{{10}}} add to {{{6}}}. So this shows us that {{{-4z+10z=6z}}}.



{{{5z^2+highlight(-4z+10z)-8}}} Replace the second term {{{6z}}} with {{{-4z+10z}}}.



{{{(5z^2-4z)+(10z-8)}}} Group the terms into two pairs.



{{{z(5z-4)+(10z-8)}}} Factor out the GCF {{{z}}} from the first group.



{{{z(5z-4)+2(5z-4)}}} 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)(5z-4)}}} Combine like terms. Or factor out the common term {{{5z-4}}}



{{{(3x+y+2)(5(3x+y)-4)}}} Plug in {{{z=3x+y}}}



{{{(3x+y+2)(15x+5y-4)}}} Distribute


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Answer:



So {{{5(3x+y)^2+6(3x+y)-8}}} factors to {{{(3x+y+2)(15x+5y-4)}}}.



In other words, {{{5(3x+y)^2+6(3x+y)-8=(3x+y+2)(15x+5y-4)}}}



Note: you can check the answer by expanding {{{(3x+y+2)(15x+5y-4)}}} to get {{{5(3x+y)^2+6(3x+y)-8}}}.