Solve the following equation by factorization.
`6x^(2)-x-2=0`

Answer

To solve the equation \(6x^2 - x - 2 = 0\) by factorization, we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \(ax^2 + bx + c = 0\), where:
- \(a = 6\)
- \(b = -1\)
- \(c = -2\)

### Step 2: Calculate the product and sum
We need to find two numbers that multiply to \(a \cdot c\) (which is \(6 \cdot -2 = -12\)) and add up to \(b\) (which is \(-1\)).

### Step 3: Find the two numbers
We are looking for two numbers that multiply to \(-12\) and add to \(-1\). The numbers are \(-4\) and \(3\) because:
- \(-4 \cdot 3 = -12\)
- \(-4 + 3 = -1\)

### Step 4: Rewrite the middle term
We can rewrite the equation by splitting the middle term using the two numbers found:
\[ 6x^2 - 4x + 3x - 2 = 0 \]

### Step 5: Factor by grouping
Now we group the terms:
\[ (6x^2 - 4x) + (3x - 2) = 0 \]
Factoring out the common terms in each group:
\[ 2x(3x - 2) + 1(3x - 2) = 0 \]

### Step 6: Factor out the common binomial
Now we can factor out the common binomial \((3x - 2)\):
\[ (3x - 2)(2x + 1) = 0 \]

### Step 7: Set each factor to zero
Now we set each factor equal to zero:
1. \(3x - 2 = 0\)
2. \(2x + 1 = 0\)

### Step 8: Solve for \(x\)
Solving the first equation:
\[ 3x - 2 = 0 \implies 3x = 2 \implies x = \frac{2}{3} \]

Solving the second equation:
\[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \]

### Final Solution
The solutions to the equation \(6x^2 - x - 2 = 0\) are:
\[ x = \frac{2}{3} \quad \text{and} \quad x = -\frac{1}{2} \]

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