Question 626283
I'm assuming you want to factor this.


Looking at the expression {{{y^2+ty-12t^2}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{1}}}, and the last coefficient is {{{-12}}}.



Now multiply the first coefficient {{{1}}} by the last coefficient {{{-12}}} to get {{{(1)(-12)=-12}}}.



Now the question is: what two whole numbers multiply to {{{-12}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{1}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-12}}} (the previous product).



Factors of {{{-12}}}:

1,2,3,4,6,12

-1,-2,-3,-4,-6,-12



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



These factors pair up and multiply to {{{-12}}}.

1*(-12) = -12
2*(-6) = -12
3*(-4) = -12
(-1)*(12) = -12
(-2)*(6) = -12
(-3)*(4) = -12


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{1}}}:



<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>-12</font></td><td  align="center"><font color=black>1+(-12)=-11</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>2+(-6)=-4</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>3+(-4)=-1</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>12</font></td><td  align="center"><font color=black>-1+12=11</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>-2+6=4</font></td></tr><tr><td  align="center"><font color=red>-3</font></td><td  align="center"><font color=red>4</font></td><td  align="center"><font color=red>-3+4=1</font></td></tr></table>



From the table, we can see that the two numbers {{{-3}}} and {{{4}}} add to {{{1}}} (the middle coefficient).



So the two numbers {{{-3}}} and {{{4}}} both multiply to {{{-12}}} <font size=4><b>and</b></font> add to {{{1}}}



Now replace the middle term {{{1ty}}} with {{{-3ty+4ty}}}. Remember, {{{-3}}} and {{{4}}} add to {{{1}}}. So this shows us that {{{-3ty+4ty=1ty}}}.



{{{y^2+highlight(-3ty+4ty)-12t^2}}} Replace the second term {{{1ty}}} with {{{-3ty+4ty}}}.



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



{{{y(y-3t)+(4ty-12t^2)}}} Factor out the GCF {{{y}}} from the first group.



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



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



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Answer:



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



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



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


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