Question 185549


Looking at the expression {{{6x^2+x-2}}}, we can see that the first coefficient is {{{6}}}, the second coefficient is {{{1}}}, and the last term is {{{-2}}}.



Now multiply the first coefficient {{{6}}} by the last term {{{-2}}} to get {{{(6)(-2)=-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)
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 {{{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 {{{1x}}} with {{{-3x+4x}}}. Remember, {{{-3}}} and {{{4}}} add to {{{1}}}. So this shows us that {{{-3x+4x=1x}}}.



{{{6x^2+highlight(-3x+4x)-2}}} Replace the second term {{{1x}}} with {{{-3x+4x}}}.



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



{{{3x(2x-1)+(4x-2)}}} Factor out the GCF {{{3x}}} from the first group.



{{{3x(2x-1)+2(2x-1)}}} 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.



{{{(3x+2)(2x-1)}}} Combine like terms. Or factor out the common term {{{2x-1}}}


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Answer:



So {{{6x^2+x-2}}} factors to {{{(3x+2)(2x-1)}}}.



This means that {{{3x+2}}} and {{{2x-1}}} are factors of {{{6x^2+x-2}}}