Question

How many and of which kind of roots does the equation have

Answer 1

The Solution:

Given:

Required:

Find the number and types of roots the function has.

$f(x)=x^3-x^2-x+1$$\begin{gathered} f(1)=1^3-(1^2)-1+1=1-1-1+1=0 \\ This\text{ means that: }x=1\text{ is a root of the function.} \end{gathered}$

Find the other root.

Factorize the function:

$\begin{gathered} f(x)=x^3-x^2-x+1=0 \\ x^2(x-1)-1(x-1)=0 \end{gathered}$$(x-1)(x^2-1)=0$$(x-1)(x-1)(x+1)=0$

So, the complete roots of the function are:

$\begin{gathered} x-1=0 \\ x=1\text{ \lparen multiplicity of 2\rparen.} \\ x+1=0 \\ x=-1 \end{gathered}$

Thus,

Answer:

The number of roots is 3. (x = 1, x = 1, and x = -1).

The roots are all real roots.

[option D]