SOLUTION: How to solve this, 6x^3-6x^2-x+1 by factoring through grouping.

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Question 392345: How to solve this, 6x^3-6x^2-x+1 by factoring through grouping.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
6x%5E3-6x%5E2-x%2B1+
%286x%5E3-6x%5E2%29-+%28x-1%29+
6x%5E2%28x-1%29-+%28x-1%29+
%28x-1%29%286x%5E2-1%29
let see if %286x%5E2-1%29
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 6x%5E2%2B0x-1, we can see that the first coefficient is 6, the second coefficient is 0, and the last term is -1.



Now multiply the first coefficient 6 by the last term -1 to get %286%29%28-1%29=-6.



Now the question is: what two whole numbers multiply to -6 (the previous product) and add to the second coefficient 0?



To find these two numbers, we need to list all 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 0:



First NumberSecond NumberSum
1-61+(-6)=-5
2-32+(-3)=-1
-16-1+6=5
-23-2+3=1




From the table, we can see that there are no pairs of numbers which add to 0. So 6x%5E2%2B0x-1 cannot be factored.



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Answer:



So 6%2Ax%5E2%2B0%2Ax-1 doesn't factor at all (over the rational numbers).



So 6%2Ax%5E2%2B0%2Ax-1 is prime.




since %286x%5E2-1%29 cannot be factored, this is your answer:%28x-1%29%286x%5E2-1%29