Question 385276


Looking at the expression {{{a^2-6ab+96b^2}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-6}}}, and the last coefficient is {{{96}}}.



Now multiply the first coefficient {{{1}}} by the last coefficient {{{96}}} to get {{{(1)(96)=96}}}.



Now the question is: what two whole numbers multiply to {{{96}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-6}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{96}}} (the previous product).



Factors of {{{96}}}:

1,2,3,4,6,8,12,16,24,32,48,96

-1,-2,-3,-4,-6,-8,-12,-16,-24,-32,-48,-96



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



These factors pair up and multiply to {{{96}}}.

1*96 = 96
2*48 = 96
3*32 = 96
4*24 = 96
6*16 = 96
8*12 = 96
(-1)*(-96) = 96
(-2)*(-48) = 96
(-3)*(-32) = 96
(-4)*(-24) = 96
(-6)*(-16) = 96
(-8)*(-12) = 96


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-6}}}:



<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>96</font></td><td  align="center"><font color=black>1+96=97</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>48</font></td><td  align="center"><font color=black>2+48=50</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>32</font></td><td  align="center"><font color=black>3+32=35</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>24</font></td><td  align="center"><font color=black>4+24=28</font></td></tr><tr><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>16</font></td><td  align="center"><font color=black>6+16=22</font></td></tr><tr><td  align="center"><font color=black>8</font></td><td  align="center"><font color=black>12</font></td><td  align="center"><font color=black>8+12=20</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-96</font></td><td  align="center"><font color=black>-1+(-96)=-97</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-48</font></td><td  align="center"><font color=black>-2+(-48)=-50</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-32</font></td><td  align="center"><font color=black>-3+(-32)=-35</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>-24</font></td><td  align="center"><font color=black>-4+(-24)=-28</font></td></tr><tr><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>-16</font></td><td  align="center"><font color=black>-6+(-16)=-22</font></td></tr><tr><td  align="center"><font color=black>-8</font></td><td  align="center"><font color=black>-12</font></td><td  align="center"><font color=black>-8+(-12)=-20</font></td></tr></table>



From the table, we can see that there are no pairs of numbers which add to {{{-6}}}. So {{{a^2-6ab+96b^2}}} cannot be factored.



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<a name="ans">


Answer:



So {{{a^2-6ab+96b^2}}} doesn't factor at all (over the rational numbers).



So {{{a^2-6ab+96b^2}}} is prime.



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Jim