`15x^(2)-7x-36=0:`

Options

A. `(5)/(9),-(4)/(3)`
B. `(9)/(5),-(4)/(3)`
C. `(9)/(5),-(3)/(4)`
D. none of these

Answer

To solve the quadratic equation \( 15x^2 - 7x - 36 = 0 \) using the factorization method, we will follow these steps:

### Step 1: Identify coefficients
The given quadratic equation is in the standard form \( ax^2 + bx + c = 0 \). Here, we have:
- \( a = 15 \)
- \( b = -7 \)
- \( c = -36 \)

### Step 2: Calculate the product \( ac \)
We need to calculate the product of \( a \) and \( c \):
\[ ac = 15 \times (-36) = -540 \]

### Step 3: Find two numbers that multiply to \( ac \) and add to \( b \)
We need to find two numbers that multiply to \( -540 \) and add up to \( -7 \). After checking the factors of \( -540 \), we find that:
\[ 27 \text{ and } -20 \text{ satisfy } 27 \times (-20) = -540 \text{ and } 27 + (-20) = 7 \]

### Step 4: Rewrite the middle term using the two numbers
We can rewrite the equation \( 15x^2 - 7x - 36 = 0 \) as:
\[ 15x^2 + 27x - 20x - 36 = 0 \]

### Step 5: Group the terms
Next, we group the terms:
\[ (15x^2 + 27x) + (-20x - 36) = 0 \]

### Step 6: Factor by grouping
Now we factor out the common factors from each group:
\[ 3x(5x + 9) - 4(5x + 9) = 0 \]
This can be factored further as:
\[ (3x - 4)(5x + 9) = 0 \]

### Step 7: Set each factor to zero
Now we set each factor equal to zero:
1. \( 3x - 4 = 0 \)
2. \( 5x + 9 = 0 \)

### Step 8: Solve for \( x \)
For the first equation:
\[ 3x - 4 = 0 \implies 3x = 4 \implies x = \frac{4}{3} \]

For the second equation:
\[ 5x + 9 = 0 \implies 5x = -9 \implies x = -\frac{9}{5} \]

### Final Solutions
Thus, the solutions to the equation \( 15x^2 - 7x - 36 = 0 \) are:
\[ x = \frac{4}{3} \quad \text{and} \quad x = -\frac{9}{5} \]

---