$$ x^{2}+2 x-8=0 $$ $$ \left(x^{2}+4 x\right)(-2 x-8)=0 $$ $$ x(x+4)-2(x+4)=0 $$ $$ (x-2)(x-4)=0 $$ $$ x-2=0 $$ or $$ x+4=0 $$ $$ x-2 $$ or $$ x=4 $$

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UpStudy AI · Accepted Answer

It looks like you've made a small mistake in the final step of your solution. Let's go through each step to identify and correct it.

### Given Equation:
$$ x^{2} + 2x - 8 = 0 $$

### Step 1: Factor the Equation
To factor the quadratic equation, we look for two numbers that multiply to **-8** (the constant term) and add up to **2** (the coefficient of the middle term).

Those numbers are **4** and **-2** because:
$$ 4 \times (-2) = -8 $$
$$ 4 + (-2) = 2 $$

So, we can rewrite the equation as:
$$ x^{2} + 4x - 2x - 8 = 0 $$

### Step 2: Group Terms
Group the terms to factor by grouping:
$$ (x^{2} + 4x) + (-2x - 8) = 0 $$
$$ x(x + 4) - 2(x + 4) = 0 $$

### Step 3: Factor Out the Common Term
Factor out the common term $$(x + 4)$$:
$$ (x + 4)(x - 2) = 0 $$

### Step 4: Apply the Zero Product Property
Set each factor equal to zero:
1. $$ x + 4 = 0 $$  
   $$ x = -4 $$

2. $$ x - 2 = 0 $$  
   $$ x = 2 $$

### Final Solutions:
$$ x = 2 \quad \text{or} \quad x = -4 $$

### Correction in Your Final Step
In your final step, you wrote:
$$ x - 2 \quad \text{or} \quad x = 4 $$

This should be corrected to:
$$ x = 2 \quad \text{or} \quad x = -4 $$

### Summary
The solutions to the equation $$ x^{2} + 2x - 8 = 0 $$ are:
$$ x = 2 \quad \text{and} \quad x = -4 $$
The solutions are $$ x = 2 $$ and $$ x = -4 $$.

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