Question 697185: 3x^2-4x+2
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! I'm assuming you want to factor.
Looking at the expression  , we can see that the first coefficient is  , the second coefficient is  , and the last term is  .
Now multiply the first coefficient by the last term to get  .
Now the question is: what two whole numbers multiply to  (the previous product) and add to the second coefficient ?
To find these two numbers, we need to list all of the factors of (the previous product).
Factors of :
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 .
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 :
| First Number | Second Number | Sum | | 1 | 6 | 1+6=7 | | 2 | 3 | 2+3=5 | | -1 | -6 | -1+(-6)=-7 | | -2 | -3 | -2+(-3)=-5 |
From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.
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Answer:
So doesn't factor at all (over the rational numbers).
So is prime.
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