Please show me how to factor this completely

x^3 - 2x^2 - 5x +6 = 0

Please show me in detail.
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A website to complement your explanation would be good to.

You can use the remainder theorem to factor out the above equation.

All you need to do is find factors x-c such that when you plug in c into the polynomial, it will give you 0.

Also c has to be a factor of the last term in the polynomial which is 6 in our case.

So c can be +/-1, +/-2 or +/-3.

Try each one.

(1)^3 - 2(1)^2 - 5(1) + 6 = 0. So x-1 is a factor

(-1)^3 - 2(-1)^2 - 5(-1) + 6 = 2.

(2)^3 - 2(2)^2 - 5(2) + 6 = -4.

(-2)^3 - 2(-2)^2 - 5(-2) + 6 = 0. x+2 is a factor

(3)^3 - 2(3)^2 - 5(3) + 6 = 0. x-3 is a factor

(-3)^3 - 2(-3)^2 - 5(-3) + 6 = -24.

Therefore, after you factor this completely, you get this:

x^3 - 2x^2 - 5x +6 = (x - 1) (x + 2) (x - 3).

Hope this helps!

x^3 - 2x^2 - 5x +6 = 0

x=1 is a solution, therefore (x-1) is a factor; so divide by this.

x-1 )x^3 - 2x^2 - 5x +6 ( x^2-x-6
___x^3 - x^2
_______-x^2 - 5x
_______-x^2 +x
___________-6x + 6
___________-6x + 6
_____________0
So factors are:
(x-1)(x^2-x-6)=(x-1)(x+2)(x-3)

(x-1)(x+2)(x-3)=0

x = 1, -2, 3

http://www.wolframalpha.com/input/?i=factor+x%5E3+-+2x%5E2+-+5x+%2B6+%3D+0