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How many x-intercepts appear on the graph of this polynomial function?

[tex]\( f(x) = x^4 - x^3 + x^2 - x \)[/tex]

A. 1 x-intercept
B. 2 x-intercepts
C. 3 x-intercepts
D. 4 x-intercepts


Sagot :

To determine the number of [tex]\( x \)[/tex]-intercepts of the polynomial function [tex]\( f(x) = x^4 - x^3 + x^2 - x \)[/tex], we need to solve for [tex]\( x \)[/tex] when [tex]\( f(x) = 0 \)[/tex].

We start by factoring the polynomial:

[tex]\[ f(x) = x^4 - x^3 + x^2 - x = x(x^3 - x^2 + x - 1) \][/tex]

Next, we need to solve the equation,

[tex]\[ x(x^3 - x^2 + x - 1) = 0 \][/tex]

This gives two cases:

1. [tex]\( x = 0 \)[/tex]

2. [tex]\( x^3 - x^2 + x - 1 = 0 \)[/tex]

For the second case, let's solve the cubic equation [tex]\( x^3 - x^2 + x - 1 = 0 \)[/tex].

The solutions to this cubic equation are:

[tex]\[ x = 1, \quad x = -i, \quad x = i \][/tex]

where [tex]\( i \)[/tex] is the imaginary unit ([tex]\( i^2 = -1 \)[/tex]).

Thus, the [tex]\( x \)[/tex]-intercepts are:

[tex]\[ 0, \, 1, \, -i, \, i \][/tex]

However, we are specifically interested in the real [tex]\( x \)[/tex]-intercepts for the graph, and only real numbers can be [tex]\( x \)[/tex]-intercepts on the Cartesian plane. From the above solutions, the real [tex]\( x \)[/tex]-intercepts are:

[tex]\[ 0, \, 1 \][/tex]

This means there are two real [tex]\( x \)[/tex]-intercepts.

Therefore, the number of [tex]\( x \)[/tex]-intercepts on the graph of the polynomial function [tex]\( f(x) = x^4 - x^3 + x^2 - x \)[/tex] are:

[tex]\[ 2 \, \text{real \( x \)-intercepts} \][/tex]

So out of the given options, the correct answer is:

[tex]\[ 2 \, \text{x-intercepts} \][/tex]