Question 674305


Looking at the expression {{{x^2-3xy+9y^2}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-3}}}, and the last coefficient is {{{9}}}.



Now multiply the first coefficient {{{1}}} by the last coefficient {{{9}}} to get {{{(1)(9)=9}}}.



Now the question is: what two whole numbers multiply to {{{9}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-3}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{9}}} (the previous product).



Factors of {{{9}}}:

1,3,9

-1,-3,-9



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



These factors pair up and multiply to {{{9}}}.

1*9 = 9
3*3 = 9
(-1)*(-9) = 9
(-3)*(-3) = 9


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-3}}}:



<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>9</font></td><td  align="center"><font color=black>1+9=10</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>3+3=6</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-9</font></td><td  align="center"><font color=black>-1+(-9)=-10</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-3+(-3)=-6</font></td></tr></table>



From the table, we can see that there are no pairs of numbers which add to {{{-3}}}. So {{{x^2-3xy+9y^2}}} cannot be factored.



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Answer:



So {{{x^2-3xy+9y^2}}} doesn't factor at all (over the rational numbers).



So {{{x^2-3xy+9y^2}}} is prime.