The number of partial fractions of `(2)/(x^(4)+x^(2)+1)` is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To determine the number of partial fractions for the expression \(\frac{2}{x^4 + x^2 + 1}\), we will follow these steps:

### Step 1: Factor the Denominator
We need to factor the polynomial in the denominator \(x^4 + x^2 + 1\).

1. Notice that \(x^4 + x^2 + 1\) can be treated as a quadratic in terms of \(x^2\).
2. Let \(y = x^2\). Then, we rewrite the expression as \(y^2 + y + 1\).
3. To factor \(y^2 + y + 1\), we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} \]
This gives us complex roots:
\[ y = \frac{-1 \pm i\sqrt{3}}{2} \]
4. Replacing \(y\) back with \(x^2\), we can express the factors as:
\[ x^2 + \frac{-1 + i\sqrt{3}}{2} \quad \text{and} \quad x^2 + \frac{-1 - i\sqrt{3}}{2} \]

### Step 2: Express in Partial Fractions
Now, we can express the fraction \(\frac{2}{x^4 + x^2 + 1}\) in terms of partial fractions. Since we have two irreducible quadratic factors, we can write:
\[ \frac{2}{x^4 + x^2 + 1} = \frac{Ax + B}{x^2 + \frac{-1 + i\sqrt{3}}{2}} + \frac{Cx + D}{x^2 + \frac{-1 - i\sqrt{3}}{2}} \]
where \(A\), \(B\), \(C\), and \(D\) are constants to be determined.

### Step 3: Count the Number of Partial Fractions
Each irreducible quadratic factor contributes one partial fraction. Since we have two irreducible quadratic factors, the total number of partial fractions is:
\[ \text{Number of Partial Fractions} = 2 \]

### Final Answer
Thus, the number of partial fractions of \(\frac{2}{x^4 + x^2 + 1}\) is **2**.

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