Question 697185
I'm assuming you want to factor.




Looking at the expression {{{3x^2-4x+2}}}, we can see that the first coefficient is {{{3}}}, the second coefficient is {{{-4}}}, and the last term is {{{2}}}.



Now multiply the first coefficient {{{3}}} by the last term {{{2}}} to get {{{(3)(2)=6}}}.



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



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{6}}} (the previous product).



Factors of {{{6}}}:

1,2,3,6

-1,-2,-3,-6



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



These factors pair up and multiply to {{{6}}}.

1*6 = 6
2*3 = 6
(-1)*(-6) = 6
(-2)*(-3) = 6


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-4}}}:



<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>6</font></td><td  align="center"><font color=black>1+6=7</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>2+3=5</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-6</font></td><td  align="center"><font color=black>-1+(-6)=-7</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-2+(-3)=-5</font></td></tr></table>



From the table, we can see that there are no pairs of numbers which add to {{{-4}}}. So {{{3x^2-4x+2}}} cannot be factored.



===============================================================


<a name="ans">


Answer:



So {{{3x^2-4x+2}}} doesn't factor at all (over the rational numbers).



So {{{3x^2-4x+2}}} is prime.