Solve the following equation: x^(3) - 6x^(2) + 11x - 6 = 0
x3 - 6x2 + 11x - 6 = 0
Let (x-a)(x-b)(x-c) = x3 - 6x2 + 11x - 6=> abc = -6 and a + b + c = -6
From abc=-6, find the possibilities of their values: (1,2,-3), (1,-2,3), (-1,2,3), (-1,-2,-3), (1,1,-6), (1,-1,6), (-1,1,6) & (-1,-1,-6)
Then find the sum of each set of brackets (making sure you do this for all of them):e.g. 1 + 2 + (-3) = 0 and (-1) + (-2) + (-3) = -6
Now, using a + b + c = -6, eliminate the values which do not work.
In this case, the only set that work are (-1,-2,-3) .
So, giving us (x-1)(x-2)(x-3) = 0
To find x solve each bracket:=> x-1=0, x-2=0 & x-3=0
=> x=1, x=2 & x=3
Hence, these are the 3 values of x.
[=> means "implies"]
EH
