The rational root theorem tells us that rational roots to a polynomial equation with integer coefficients can be written in the form p/qpq, where pp is a factor of the constant term and qq is a factor of the leading coefficient.
The polynomial equation is 1*x^3 - 8x^2 + 25x - 26 = 01⋅x3−8x2+25x−26=0.
The integer factors of the constant -26−26 are +-26, +-13,+-2,+-1±26,±13,±2,±1.
The integer factors of the leading coefficient 11 are +-1±1.
The rational solutions must therefore be among the following: +-26/1, +-13/1, +-2/1, +-1/1±261,±131,±21,±11
Try each answer by substituting them into xx in the equation until you find the solution. You will find that x = 2x=2 solves the equation. This is the first solution.
Now we know that (x-2)(x−2) is a factor of the polynomial, meaning that x^3 - 8x^2 + 25x - 26 = (x-2)*g(x)x3−8x2+25x−26=(x−2)⋅g(x) where g(x)g(x) is some polynomial of a lower degree. Use polynomial division to find:
g(x) = (x^3 - 8x^2 + 25x - 26) / (x-2)=x^2−6x+13 g(x)=x3−8x2+25x−26x−2=x2−6x+13
(Polynomial divison might need its own explanation for those who haven't learned it. If there's a separate Socratic thread on it, it should be linked here.)
With the factorized polynomial (x-2)(x^2−6x+13) = 0(x−2)(x2−6x+13)=0, now solve for xx when (x^2−6x+13) = 0(x2−6x+13)=0
A quadratic equation ax^2+bx+c=0ax2+bx+c=0 can be solved using the quadratic formula: x = (-b+- sqrt(b^2 -4ac))/(2a)x=−b±√b2−4ac2a.
Plug a = 1a=1, b =-6b=−6 and c = 13c=13 into it and we get:
x = 3+-sqrt(-4) x=3±√−4
= 3+-sqrt(4*-1) =3±√4⋅−1
= 3+-sqrt(4)sqrt(-1) =3±√4√−1
= 3+-2i=3±2i
Now we have found all three solutions:
x_0 = 2x0=2
x_1 = 3+2ix1=3+2i
x_2 = 3-2ix2=3−2i
