How do you find all zeros of #F(x)= x^3 -8x^2 +25x -26#?

1 Answer
Sep 19, 2015

Answer:

There are three solutions:
#x_0 = 2#
#x_1 = 3+2i#
#x_2 = 3-2i#

Explanation:

The rational root theorem tells us that rational roots to a polynomial equation with integer coefficients can be written in the form #p/q#, where #p# is a factor of the constant term and #q# is a factor of the leading coefficient.

The polynomial equation is #1*x^3 - 8x^2 + 25x - 26 = 0#.
The integer factors of the constant #-26# are #+-26, +-13,+-2,+-1#.
The integer factors of the leading coefficient #1# are #+-1#.

The rational solutions must therefore be among the following: #+-26/1, +-13/1, +-2/1, +-1/1#

Try each answer by substituting them into #x# in the equation until you find the solution. You will find that #x = 2# solves the equation. This is the first solution.

Now we know that #(x-2)# is a factor of the polynomial, meaning that #x^3 - 8x^2 + 25x - 26 = (x-2)*g(x)# where #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 #

(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#, now solve for #x# when #(x^2−6x+13) = 0#

A quadratic equation #ax^2+bx+c=0# can be solved using the quadratic formula: #x = (-b+- sqrt(b^2 -4ac))/(2a)#.

Plug #a = 1#, #b =-6# and #c = 13# into it and we get:

#x = 3+-sqrt(-4) #
#= 3+-sqrt(4*-1) #
#= 3+-sqrt(4)sqrt(-1) #
#= 3+-2i#

Now we have found all three solutions:
#x_0 = 2#
#x_1 = 3+2i#
#x_2 = 3-2i#