To solve \( 6x^2 - x - 2 = 0 \) using the area model, we can use factoring by splitting the middle term. Here's a step-by-step guide on how to do this:

### Step 1: Multiply the coefficient of \(x^2\) (6) and the constant term (-2).
We have:
\[
6 \times (-2) = -12
\]

### Step 2: Find two numbers that multiply to give -12 and add to give the middle coefficient (-1).
These numbers are **-4** and **3**, because:
\[
-4 \times 3 = -12 \quad \text{and} \quad -4 + 3 = -1
\]

### Step 3: Rewrite the middle term using these two numbers.
So, the equation becomes:
\[
6x^2 - 4x + 3x - 2 = 0
\]

### Step 4: Group the terms and factor each group.
Group the terms as follows:
\[
(6x^2 - 4x) + (3x - 2) = 0
\]

Now factor each group:
\[
2x(3x - 2) + 1(3x - 2) = 0
\]

### Step 5: Factor out the common binomial factor.
Both groups contain the factor \( (3x - 2) \), so we factor that out:
\[
(2x + 1)(3x - 2) = 0
\]

### Step 6: Solve for \(x\).
Set each factor equal to zero:
1. \( 2x + 1 = 0 \) gives \( x = -\frac{1}{2} \)
2. \( 3x - 2 = 0 \) gives \( x = \frac{2}{3} \)

So the solutions are:
\[
x = -\frac{1}{2} \quad \text{or} \quad x = \frac{2}{3}
\]

---

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:
1. How does the area model help with factoring quadratic equations?
2. Can you factor other quadratic equations using the same method?
3. How would the area model work for a quadratic that doesn’t factor easily?
4. What is the role of the discriminant in solving quadratic equations?
5. How can you check if your factored form is correct?

**Tip:** Always check your solutions by substituting them back into the original equation!

Discover how to solve the quadratic equation 6x^2 - x - 2 using the area model. This step-by-step guide explains factoring techniques and key algebraic concepts. Ideal for students in Grades 9-10.

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