x^3  2x^2  5x +6 = 0
Please show me in detail.
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Please show me in detail.
Best answer will be selected today.
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 xc 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 x1 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. x3 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!
All you need to do is find factors xc 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 x1 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. x3 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 (x1) is a factor; so divide by this.
x1 )x^3  2x^2  5x +6 ( x^2x6
___x^3  x^2
_______x^2  5x
_______x^2 +x
___________6x + 6
___________6x + 6
_____________0
So factors are:
(x1)(x^2x6)=(x1)(x+2)(x3)
(x1)(x+2)(x3)=0
x = 1, 2, 3
x=1 is a solution, therefore (x1) is a factor; so divide by this.
x1 )x^3  2x^2  5x +6 ( x^2x6
___x^3  x^2
_______x^2  5x
_______x^2 +x
___________6x + 6
___________6x + 6
_____________0
So factors are:
(x1)(x^2x6)=(x1)(x+2)(x3)
(x1)(x+2)(x3)=0
x = 1, 2, 3

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