Mathematics -- solving higher order equation of one variable The third power of X - the square of 3x + X + 2 = 0 Let f = (x) = the third power of X - the square of 3x + X + 2 ∵f(2)=8-3*4+2+2=0 X-2 is a factor of the square + X + 2 of the cubic power - 3x of polynomial X (I understand here) The original equation can be solved as: (X-2) (the square of x-x-1) = 0 (I want to know how to push this step out and how to get the result at once. I hope you can give me a detailed and understandable answer.)

Mathematics -- solving higher order equation of one variable The third power of X - the square of 3x + X + 2 = 0 Let f = (x) = the third power of X - the square of 3x + X + 2 ∵f(2)=8-3*4+2+2=0 X-2 is a factor of the square + X + 2 of the cubic power - 3x of polynomial X (I understand here) The original equation can be solved as: (X-2) (the square of x-x-1) = 0 (I want to know how to push this step out and how to get the result at once. I hope you can give me a detailed and understandable answer.)

It should be ax ^ 2 + BX + C
Because the coefficient of x ^ 3 is 1, a = 1
Constant term: C = 2 / (- 2) = - 1
The term x ^ 2 is: - 2A + B = - 3, so B = - 1
That is x ^ 2-x-1