Question 681794


Looking at the expression {{{40a^2+21a-2}}}, we can see that the first coefficient is {{{40}}}, the second coefficient is {{{21}}}, and the last term is {{{-2}}}.



Now multiply the first coefficient {{{40}}} by the last term {{{-2}}} to get {{{(40)(-2)=-80}}}.



Now the question is: what two whole numbers multiply to {{{-80}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{21}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{-80}}} (the previous product).



Factors of {{{-80}}}:

1,2,4,5,8,10,16,20,40,80

-1,-2,-4,-5,-8,-10,-16,-20,-40,-80



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



These factors pair up and multiply to {{{-80}}}.

1*(-80) = -80
2*(-40) = -80
4*(-20) = -80
5*(-16) = -80
8*(-10) = -80
(-1)*(80) = -80
(-2)*(40) = -80
(-4)*(20) = -80
(-5)*(16) = -80
(-8)*(10) = -80


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{21}}}:



<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>-80</font></td><td  align="center"><font color=black>1+(-80)=-79</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>-40</font></td><td  align="center"><font color=black>2+(-40)=-38</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>-20</font></td><td  align="center"><font color=black>4+(-20)=-16</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>-16</font></td><td  align="center"><font color=black>5+(-16)=-11</font></td></tr><tr><td  align="center"><font color=black>8</font></td><td  align="center"><font color=black>-10</font></td><td  align="center"><font color=black>8+(-10)=-2</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>80</font></td><td  align="center"><font color=black>-1+80=79</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>40</font></td><td  align="center"><font color=black>-2+40=38</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>20</font></td><td  align="center"><font color=black>-4+20=16</font></td></tr><tr><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>16</font></td><td  align="center"><font color=black>-5+16=11</font></td></tr><tr><td  align="center"><font color=black>-8</font></td><td  align="center"><font color=black>10</font></td><td  align="center"><font color=black>-8+10=2</font></td></tr></table>



From the table, we can see that there are no pairs of numbers which add to {{{21}}}. So {{{40a^2+21a-2}}} cannot be factored.



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Answer:



So {{{40a^2+21a-2}}} doesn't factor at all (over the rational numbers).



So {{{40a^2+21a-2}}} is prime.