In how many different points does the graph of the function `f(x)=x^(3)-2x^(2)+x-2` intersect the x-axis?

Options

A. `0`
B. `1`
C. `2`
D. `3`

Answer

To determine how many different points the graph of the function \( f(x) = x^3 - 2x^2 + x - 2 \) intersects the x-axis, we need to find the roots of the equation \( f(x) = 0 \). This means we need to solve the equation:

\[ x^3 - 2x^2 + x - 2 = 0 \]

### Step 1: Factor the polynomial
We can try to factor the polynomial. First, we can look for rational roots using the Rational Root Theorem. The possible rational roots are the factors of the constant term (-2) divided by the factors of the leading coefficient (1). The possible rational roots are \( \pm 1, \pm 2 \).

### Step 2: Test possible roots
Let's test \( x = 2 \):

\[ f(2) = 2^3 - 2(2^2) + 2 - 2 = 8 - 8 + 2 - 2 = 0 \]

Since \( f(2) = 0 \), \( x = 2 \) is a root.

### Step 3: Perform polynomial long division
Now that we know \( x - 2 \) is a factor, we can divide the polynomial \( f(x) \) by \( x - 2 \) to find the other factors.

Using synthetic division:

\[ \begin{array}{r|rrrr} 2 & 1 & -2 & 1 & -2 \\ & & 2 & 0 & 2 \\ \hline & 1 & 0 & 1 & 0 \\ \end{array} \]

The result is \( x^2 + 1 \), so we can write:

\[ f(x) = (x - 2)(x^2 + 1) \]

### Step 4: Solve the remaining factor
Now we need to solve \( x^2 + 1 = 0 \):

\[ x^2 = -1 \]

This gives us:

\[ x = \pm i \]

### Step 5: Determine the number of real intersections
The only real solution we have is \( x = 2 \). The solutions \( x = i \) and \( x = -i \) are imaginary and do not correspond to points where the graph intersects the x-axis.

### Conclusion
Thus, the graph of the function \( f(x) = x^3 - 2x^2 + x - 2 \) intersects the x-axis at only one point, which is at \( x = 2 \).

The answer is:

**The graph intersects the x-axis at 1 point.**

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