Find the values of x for which `f(x)=(x^(2)+7)/(x^(3)+3x^(2)-x-3)` is discontinuos.

Answer

To find the values of \( x \) for which the function

\[ f(x) = \frac{x^2 + 7}{x^3 + 3x^2 - x - 3} \]

is discontinuous, we need to determine when the denominator is equal to zero, as a function is discontinuous at points where its denominator is zero.

### Step 1: Set the denominator to zero

We start by setting the denominator equal to zero:

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

### Step 2: Factor the polynomial

To solve the cubic equation, we can use the Rational Root Theorem or trial and error to find a root. Let's test \( x = 1 \):

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

Since \( x = 1 \) is a root, we can factor the polynomial by dividing it by \( x - 1 \).

### Step 3: Perform polynomial long division

We divide \( x^3 + 3x^2 - x - 3 \) by \( x - 1 \):

1. Divide the leading term: \( x^3 \div x = x^2 \)
2. Multiply \( x^2 \) by \( x - 1 \): \( x^3 - x^2 \)
3. Subtract: \( (3x^2 - (-x^2)) - (-x) - 3 = 4x^2 - x - 3 \)
4. Repeat: \( 4x^2 \div x = 4x \)
5. Multiply: \( 4x(x - 1) = 4x^2 - 4x \)
6. Subtract: \( (-x + 4x) - 3 = 3x - 3 \)
7. Repeat: \( 3x \div x = 3 \)
8. Multiply: \( 3(x - 1) = 3x - 3 \)
9. Subtract: \( 0 \)

Thus, we have:

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

### Step 4: Factor the quadratic

Next, we need to factor \( x^2 + 4x + 3 \):

\[ x^2 + 4x + 3 = (x + 1)(x + 3) \]

### Step 5: Write the complete factorization

Putting it all together, we have:

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

### Step 6: Find the points of discontinuity

The function \( f(x) \) is discontinuous where the denominator is zero. Therefore, we set each factor to zero:

1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( x + 1 = 0 \) → \( x = -1 \)
3. \( x + 3 = 0 \) → \( x = -3 \)

### Conclusion

The values of \( x \) for which \( f(x) \) is discontinuous are:

\[ x = 1, -1, -3 \]