Question 751412


Looking at the expression {{{a^2-2ab+5}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-2}}}, and the last coefficient is {{{5}}}.



Now multiply the first coefficient {{{1}}} by the last coefficient {{{5}}} to get {{{(1)(5)=5}}}.



Now the question is: what two whole numbers multiply to {{{5}}} (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 {{{5}}} (the previous product).



Factors of {{{5}}}:

1,5

-1,-5



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



These factors pair up and multiply to {{{5}}}.

1*5 = 5
(-1)*(-5) = 5


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>5</font></td><td  align="center"><font color=black>1+5=6</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>-1+(-5)=-6</font></td></tr></table>



From the table, we can see that there are no pairs of numbers which add to {{{-2}}}. So {{{a^2-2ab+5}}} cannot be factored.



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Answer:



So {{{a^2-2ab+5}}} doesn't factor at all (over the rational numbers).



So {{{a^2-2ab+5}}} is prime.



Note: there may be a typo, so make sure everything is typed correctly, thanks.