Question 870601


Looking at the expression {{{x^2-12x-5}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-12}}}, and the last term is {{{-5}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{-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 {{{-12}}}?



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 {{{-12}}}:



<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)=-4</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=4</font></td></tr></table>



From the table, we can see that there are no pairs of numbers which add to {{{-12}}}. So {{{x^2-12x-5}}} cannot be factored.



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Answer:



So {{{x^2-12x-5}}} doesn't factor at all (over the rational numbers).



So {{{x^2-12x-5}}} is prime.



If you want to solve {{{x^2-12x-5=0}}}, you have to either complete the square or use the quadratic formula.