Question 380373


Looking at the expression {{{21x^2+2x-3}}}, we can see that the first coefficient is {{{21}}}, the second coefficient is {{{2}}}, and the last term is {{{-3}}}.



Now multiply the first coefficient {{{21}}} by the last term {{{-3}}} to get {{{(21)(-3)=-63}}}.



Now the question is: what two whole numbers multiply to {{{-63}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{2}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-63}}} (the previous product).



Factors of {{{-63}}}:

1,3,7,9,21,63

-1,-3,-7,-9,-21,-63



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



These factors pair up and multiply to {{{-63}}}.

1*(-63) = -63
3*(-21) = -63
7*(-9) = -63
(-1)*(63) = -63
(-3)*(21) = -63
(-7)*(9) = -63


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{2}}}:



<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>-63</font></td><td  align="center"><font color=black>1+(-63)=-62</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>-21</font></td><td  align="center"><font color=black>3+(-21)=-18</font></td></tr><tr><td  align="center"><font color=black>7</font></td><td  align="center"><font color=black>-9</font></td><td  align="center"><font color=black>7+(-9)=-2</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>63</font></td><td  align="center"><font color=black>-1+63=62</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>21</font></td><td  align="center"><font color=black>-3+21=18</font></td></tr><tr><td  align="center"><font color=red>-7</font></td><td  align="center"><font color=red>9</font></td><td  align="center"><font color=red>-7+9=2</font></td></tr></table>



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



So the two numbers {{{-7}}} and {{{9}}} both multiply to {{{-63}}} <font size=4><b>and</b></font> add to {{{2}}}



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



{{{21x^2+highlight(-7x+9x)-3}}} Replace the second term {{{2x}}} with {{{-7x+9x}}}.



{{{(21x^2-7x)+(9x-3)}}} Group the terms into two pairs.



{{{7x(3x-1)+(9x-3)}}} Factor out the GCF {{{7x}}} from the first group.



{{{7x(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.



{{{(7x+3)(3x-1)}}} Combine like terms. Or factor out the common term {{{3x-1}}}



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Answer:



So {{{21x^2+2x-3}}} factors to {{{(7x+3)(3x-1)}}}.



In other words, {{{21x^2+2x-3=(7x+3)(3x-1)}}}.



Note: you can check the answer by expanding {{{(7x+3)(3x-1)}}} to get {{{21x^2+2x-3}}} or by graphing the original expression and the answer (the two graphs should be identical).



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Jim