Question

simpilfy x^2 -8x-9/ x^2 +6x +5
The simplified form is




What are the restrictions on the variable?
(Type only the numbers and separate your answers with commas, as needed.)

Answer 1

To simplify the expression \(\frac{x^2 - 8x - 9}{x^2 + 6x + 5}\), we can start by factoring both the numerator and the denominator.

**Step 1: Factor the numerator and the denominator.**

1. **Numerator: \(x^2 - 8x - 9\)**
To factor this, we need two numbers that multiply to \(-9\) and add up to \(-8\). The factors are \(-9\) and \(1\).
\[
x^2 - 8x - 9 = (x - 9)(x + 1)
\]

2. **Denominator: \(x^2 + 6x + 5\)**
For the denominator, we need two numbers that multiply to \(5\) and add up to \(6\). The factors are \(1\) and \(5\).
\[
x^2 + 6x + 5 = (x + 1)(x + 5)
\]

**Step 2: Rewrite the fraction with factored terms.**
\[
\frac{x^2 - 8x - 9}{x^2 + 6x + 5} = \frac{(x - 9)(x + 1)}{(x + 1)(x + 5)}
\]

**Step 3: Cancel common factors.**
We see that \(x + 1\) is common in both the numerator and the denominator:
\[
\frac{(x - 9)\cancel{(x + 1)}}{\cancel{(x + 1)}(x + 5)} = \frac{x - 9}{x + 5}
\]

**Simplified Expression:**
\[
\frac{x - 9}{x + 5}
\]

**Step 4: Determine restrictions on the variable.**
Restrictions come from values that make the denominator zero. We set the original denominator equal to zero and solve:
\[
x^2 + 6x + 5 = 0
\]
Factoring, we get:
\[
(x + 1)(x + 5) = 0 \implies x + 1 = 0 \quad \text{or} \quad x + 5 = 0
\]
This gives us:
\[
x = -1 \quad \text{and} \quad x = -5
\]

Thus, the restrictions on the variable are:
\[
-1, -5
\]

**Final Result:**
The simplified form is \(\frac{x - 9}{x + 5}\) and the restrictions on the variable are \( -1, -5\).