Question 183249
I'm assuming that you want to factor this.




Looking at the expression {{{3x^2-4x-4}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{-4}}}, and the last term is {{{-4}}}.



Now multiply the first coefficient {{{3}}} by the last term {{{-4}}} to get {{{(3)(-4)=-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 {{{-4}}}?



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


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-4}}}:



<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=red>2</font></td><td  align="center"><font color=red>-6</font></td><td  align="center"><font color=red>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=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></table>



From the table, we can see that the two numbers {{{2}}} and {{{-6}}} add to {{{-4}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{-6}}} both multiply to {{{-12}}} <font size=4><b>and</b></font> add to {{{-4}}}



Now replace the middle term {{{-4x}}} with {{{2x-6x}}}. Remember, {{{2}}} and {{{-6}}} add to {{{-4}}}. So this shows us that {{{2x-6x=-4x}}}.



{{{3x^2+highlight(2x-6x)-4}}} Replace the second term {{{-4x}}} with {{{2x-6x}}}.



{{{(3x^2+2x)+(-6x-4)}}} Group the terms into two pairs.



{{{x(3x+2)+(-6x-4)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(3x+2)-2(3x+2)}}} 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.



{{{(x-2)(3x+2)}}} Combine like terms. Or factor out the common term {{{3x+2}}}


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Answer:



So {{{3x^2-4x-4}}} factors to {{{(x-2)(3x+2)}}}.



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