Question 190939
I'm assuming you want to factor?



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



Now multiply the first coefficient {{{3}}} by the last term {{{-1}}} to get {{{(3)(-1)=-3}}}.



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



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-3}}} (the previous product).



Factors of {{{-3}}}:

1,3

-1,-3



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



These factors pair up and multiply to {{{-3}}}.

1*(-3)
(-1)*(3)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{2}}}:



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



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



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



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



{{{3n^2+highlight(-n+3n)-1}}} Replace the second term {{{2n}}} with {{{-n+3n}}}.



{{{(3n^2-n)+(3n-1)}}} Group the terms into two pairs.



{{{n(3n-1)+(3n-1)}}} Factor out the GCF {{{n}}} from the first group.



{{{n(3n-1)+1(3n-1)}}} Factor out {{{1}}} 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.



{{{(n+1)(3n-1)}}} Combine like terms. Or factor out the common term {{{3n-1}}}


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Answer:



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



Note: you can check the answer by FOILing {{{(n+1)(3n-1)}}} to get {{{3n^2+2n-1}}} or by graphing the original expression and the answer (the two graphs should be identical).