document.write( "Question 392345: How to solve this, 6x^3-6x^2-x+1 by factoring through grouping. \n" ); document.write( "

Algebra.Com's Answer #278470 by MathLover1(20849) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"6x%5E3-6x%5E2-x%2B1+\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"%286x%5E3-6x%5E2%29-+%28x-1%29+\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"6x%5E2%28x-1%29-+%28x-1%29+\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"%28x-1%29%286x%5E2-1%29\"$ \r
\n" ); document.write( "\n" ); document.write( "let see if $\"%286x%5E2-1%29\"$ \r
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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 $\"0\"$ :

\n" ); document.write( "

First Number	Second Number	Sum
1	-6	1+(-6)=-5
2	-3	2+(-3)=-1
-1	6	-1+6=5
-2	3	-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.

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

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.

\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "since $\"%286x%5E2-1%29\"$ cannot be factored, this is your answer: $\"%28x-1%29%286x%5E2-1%29\"$ \n" ); document.write( "

\n" );