While solving $$ (x-5)(x+2)=-6 $$, Kyle decides that $$ x $$ must equal 5 or -2 . "Not so fast!" exclaims Stanton. "The product does not equal zero. We need to change the equation first." a. What is Stanton talking about? b. How can the equation be rewritten? Use your algebraic tools to rewrite the equation so that it can be solved. Colve the resulting equation from part (b) for $$ x $$. Do your solutions match Kyle's?

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Let's address each part of the problem step by step.

---

### **a. What is Stanton talking about?**

Stanton is highlighting a critical oversight in Kyle's approach. Kyle assumed that for the equation $$(x - 5)(x + 2) = -6$$, setting each factor equal to zero would yield the solutions. This method—known as the **Zero Product Property**—is only valid when the product of factors equals **zero**.

However, in the given equation, the product $$(x - 5)(x + 2)$$ is equal to **$$-6$$**, not zero. Therefore, directly setting each factor to zero does not solve the original equation. Stanton is emphasizing the need to first adjust the equation so that it fits the conditions where the Zero Product Property can be appropriately applied.

---

### **b. How can the equation be rewritten?**

To solve the equation $$(x - 5)(x + 2) = -6$$ correctly, we need to rearrange it so that one side of the equation equals zero. This setup allows us to apply the Zero Product Property or other solving methods effectively. Here's how you can do it:

1. **Start with the original equation:**
   
   $$
   (x - 5)(x + 2) = -6
   $$

2. **Move $$-6$$ to the left side to set the equation to zero:**
   
   $$
   (x - 5)(x + 2) + 6 = 0
   $$

3. **Alternatively, expand the left side and combine like terms:**
   
   $$
   x^2 + 2x - 5x - 10 = -6
   $$
   
   $$
   x^2 - 3x - 10 = -6
   $$
   
   $$
   x^2 - 3x - 4 = 0
   $$

Both approaches lead to the same quadratic equation:

$$
x^2 - 3x - 4 = 0
$$

---

### **c. Solve the resulting equation from part (b) for $$x$$. Do your solutions match Kyle's?**

Now, let's solve the quadratic equation:

$$
x^2 - 3x - 4 = 0
$$

**Method 1: Factoring**

We look for two numbers that multiply to **$$-4$$** and add to **$$-3$$**. These numbers are **$$-4$$** and **$$1$$**.

$$
x^2 - 4x + x - 4 = 0
$$
$$
(x^2 - 4x) + (x - 4) = 0
$$
$$
x(x - 4) + 1(x - 4) = 0
$$
$$
(x + 1)(x - 4) = 0
$$

Setting each factor equal to zero:

$$
x + 1 = 0 \quad \Rightarrow \quad x = -1
$$
$$
x - 4 = 0 \quad \Rightarrow \quad x = 4
$$

**Method 2: Quadratic Formula**

For the equation $$x^2 - 3x - 4 = 0$$, the quadratic formula is:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

Where $$a = 1$$, $$b = -3$$, and $$c = -4$$.

$$
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)}
$$
$$
x = \frac{3 \pm \sqrt{9 + 16}}{2}
$$
$$
x = \frac{3 \pm \sqrt{25}}{2}
$$
$$
x = \frac{3 \pm 5}{2}
$$

This yields two solutions:

$$
x = \frac{3 + 5}{2} = \frac{8}{2} = 4
$$
$$
x = \frac{3 - 5}{2} = \frac{-2}{2} = -1
$$

**Comparison with Kyle's Solutions:**

- **Kyle's Solutions:** $$x = 5$$ and $$x = -2$$
- **Correct Solutions:** $$x = 4$$ and $$x = -1$$

Stanton was correct in pointing out that Kyle's solutions do not satisfy the original equation. The proper solutions after correctly rewriting and solving the equation are $$x = 4$$ and $$x = -1$$, which **do not match** Kyle's initial answers.

---

**Verification:**

To ensure the correct solutions satisfy the original equation $$(x - 5)(x + 2) = -6$$:

1. **For $$x = 4$$:**
   
   $$
   (4 - 5)(4 + 2) = (-1)(6) = -6 \quad \checkmark
   $$

2. **For $$x = -1$$:**
   
   $$
   (-1 - 5)(-1 + 2) = (-6)(1) = -6 \quad \checkmark
   $$

Both solutions satisfy the original equation, confirming their validity.
Stanton is pointing out that Kyle incorrectly assumed the Zero Product Property applies when the product equals $$-6$$ instead of zero. To solve $$(x - 5)(x + 2) = -6$$, the equation should first be rewritten as $$x^2 - 3x - 4 = 0$$. Solving this quadratic equation gives $$x = 4$$ and $$x = -1$$, which do not match Kyle's solutions of $$x = 5$$ and $$x = -2$$.

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