Question 593528
I'm assuming you want to factor this.



Looking at the expression {{{6x^2+7x-3}}}, we can see that the first coefficient is {{{6}}}, the second coefficient is {{{7}}}, and the last term is {{{-3}}}.



Now multiply the first coefficient {{{6}}} by the last term {{{-3}}} to get {{{(6)(-3)=-18}}}.



Now the question is: what two whole numbers multiply to {{{-18}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{7}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-18}}} (the previous product).



Factors of {{{-18}}}:

1,2,3,6,9,18

-1,-2,-3,-6,-9,-18



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



These factors pair up and multiply to {{{-18}}}.

1*(-18) = -18
2*(-9) = -18
3*(-6) = -18
(-1)*(18) = -18
(-2)*(9) = -18
(-3)*(6) = -18


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{7}}}:



<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>-18</font></td><td  align="center"><font color=black>1+(-18)=-17</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>-9</font></td><td  align="center"><font color=black>2+(-9)=-7</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>3+(-6)=-3</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>18</font></td><td  align="center"><font color=black>-1+18=17</font></td></tr><tr><td  align="center"><font color=red>-2</font></td><td  align="center"><font color=red>9</font></td><td  align="center"><font color=red>-2+9=7</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>-3+6=3</font></td></tr></table>



From the table, we can see that the two numbers {{{-2}}} and {{{9}}} add to {{{7}}} (the middle coefficient).



So the two numbers {{{-2}}} and {{{9}}} both multiply to {{{-18}}} <font size=4><b>and</b></font> add to {{{7}}}



Now replace the middle term {{{7x}}} with {{{-2x+9x}}}. Remember, {{{-2}}} and {{{9}}} add to {{{7}}}. So this shows us that {{{-2x+9x=7x}}}.



{{{6x^2+highlight(-2x+9x)-3}}} Replace the second term {{{7x}}} with {{{-2x+9x}}}.



{{{(6x^2-2x)+(9x-3)}}} Group the terms into two pairs.



{{{2x(3x-1)+(9x-3)}}} Factor out the GCF {{{2x}}} from the first group.



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



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



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Answer:



So {{{6x^2+7x-3}}} factors to {{{(2x+3)(3x-1)}}}.



In other words, {{{6x^2+7x-3=(2x+3)(3x-1)}}}.



Note: you can check the answer by expanding {{{(2x+3)(3x-1)}}} to get {{{6x^2+7x-3}}} or by graphing the original expression and the answer (the two graphs should be identical).


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