The easiest way to find X from this point in my calculation:
((-2-x)(3-x)(5-x)-28)-(12(3-x)+4(-2-x)-4(5-x))

I know the solutions are x=3,2,1 but how do I find X from the above, by factoring.

Question

The easiest way to find X from this point in my calculation:
((-2-x)(3-x)(5-x)-28)-(12(3-x)+4(-2-x)-4(5-x))

I know the solutions are x=3,2,1 but how do I find X from the above, by factoring.

anonymous · Answer

What class is this for?

anonymous · Answer

Algebra 1?

anonymous · Answer

I am working out the eigenvalues of a matrix and this is part of the working.

anonymous · Answer

Oh, I somewhat remember these...If I remember correctly, you have to distribute to obtain the most simple form, then you have to do synthetic division to fins the solutions.

anonymous · Answer

Does that sound right?

anonymous · Answer

I am not sure. When I expand it all it's an absolute mess but the tutor did a very simple factorisation method which I am trying to replicate :(

anonymous · Answer

Are you sure you have the right equation because the roots of that one seem to different from 1,2 and 3, but i may be wrong.

anonymous · Answer

The matrix is -2 2 3
                     -2 3 2
                     -4 2 5
I then subtract lambda so it becomes -2-x for eg. (but I will say x as I am not sure of how to type the symbol). So once I do this and use co-factor expansion I get the above equation to solve for 0. Unless I made an error somewhere :(