Question 385262
I'm assuming you want to factor this.



Looking at the expression {{{12y^2+5y-3}}}, we can see that the first coefficient is {{{12}}}, the second coefficient is {{{5}}}, and the last term is {{{-3}}}.



Now multiply the first coefficient {{{12}}} by the last term {{{-3}}} to get {{{(12)(-3)=-36}}}.



Now the question is: what two whole numbers multiply to {{{-36}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{5}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-36}}} (the previous product).



Factors of {{{-36}}}:

1,2,3,4,6,9,12,18,36

-1,-2,-3,-4,-6,-9,-12,-18,-36



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



These factors pair up and multiply to {{{-36}}}.

1*(-36) = -36
2*(-18) = -36
3*(-12) = -36
4*(-9) = -36
6*(-6) = -36
(-1)*(36) = -36
(-2)*(18) = -36
(-3)*(12) = -36
(-4)*(9) = -36
(-6)*(6) = -36


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{5}}}:



<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>-36</font></td><td  align="center"><font color=black>1+(-36)=-35</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>-18</font></td><td  align="center"><font color=black>2+(-18)=-16</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>-12</font></td><td  align="center"><font color=black>3+(-12)=-9</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>-9</font></td><td  align="center"><font color=black>4+(-9)=-5</font></td></tr><tr><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>6+(-6)=0</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>36</font></td><td  align="center"><font color=black>-1+36=35</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>18</font></td><td  align="center"><font color=black>-2+18=16</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>12</font></td><td  align="center"><font color=black>-3+12=9</font></td></tr><tr><td  align="center"><font color=red>-4</font></td><td  align="center"><font color=red>9</font></td><td  align="center"><font color=red>-4+9=5</font></td></tr><tr><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>-6+6=0</font></td></tr></table>



From the table, we can see that the two numbers {{{-4}}} and {{{9}}} add to {{{5}}} (the middle coefficient).



So the two numbers {{{-4}}} and {{{9}}} both multiply to {{{-36}}} <font size=4><b>and</b></font> add to {{{5}}}



Now replace the middle term {{{5y}}} with {{{-4y+9y}}}. Remember, {{{-4}}} and {{{9}}} add to {{{5}}}. So this shows us that {{{-4y+9y=5y}}}.



{{{12y^2+highlight(-4y+9y)-3}}} Replace the second term {{{5y}}} with {{{-4y+9y}}}.



{{{(12y^2-4y)+(9y-3)}}} Group the terms into two pairs.



{{{4y(3y-1)+(9y-3)}}} Factor out the GCF {{{4y}}} from the first group.



{{{4y(3y-1)+3(3y-1)}}} Factor out {{{3}}} 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.



{{{(4y+3)(3y-1)}}} Combine like terms. Or factor out the common term {{{3y-1}}}



===============================================================



Answer:



So {{{12y^2+5y-3}}} factors to {{{(4y+3)(3y-1)}}}.



In other words, {{{12y^2+5y-3=(4y+3)(3y-1)}}}.



Note: you can check the answer by expanding {{{(4y+3)(3y-1)}}} to get {{{12y^2+5y-3}}} or by graphing the original expression and the answer (the two graphs should be identical).



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Jim