`6y^(2)-7y-5`=________

Options

A. `(3y+5)(2y-1)`
B. `(3y-5)(2y+1)`
C. `(3y-5)(2y-1)`
D. `(3y+5)(2y+1)`

Answer

To factorize the expression \(6y^2 - 7y - 5\), we can follow these steps:

### Step 1: Identify the coefficients
The expression is in the standard quadratic form \(ay^2 + by + c\), where:
- \(a = 6\)
- \(b = -7\)
- \(c = -5\)

### Step 2: Multiply \(a\) and \(c\)
We need to multiply \(a\) and \(c\):
\[ a \cdot c = 6 \cdot (-5) = -30 \]

### Step 3: Find two numbers that multiply to \(-30\) and add to \(-7\)
We need to find two numbers that multiply to \(-30\) (the result from step 2) and add to \(-7\) (the coefficient \(b\)):
The numbers are \(-10\) and \(3\) because:
\[ -10 \cdot 3 = -30 \quad \text{and} \quad -10 + 3 = -7 \]

### Step 4: Rewrite the middle term
Now we can rewrite the expression \(6y^2 - 7y - 5\) by breaking the middle term \(-7y\) into \(-10y + 3y\):
\[ 6y^2 - 10y + 3y - 5 \]

### Step 5: Group the terms
Next, we group the terms:
\[ (6y^2 - 10y) + (3y - 5) \]

### Step 6: Factor out the common factors
Now, we factor out the common factors from each group:
- From the first group \(6y^2 - 10y\), we can factor out \(2y\):
\[ 2y(3y - 5) \]
- From the second group \(3y - 5\), we can factor out \(1\):
\[ 1(3y - 5) \]

### Step 7: Combine the factors
Now we can combine the factored terms:
\[ 2y(3y - 5) + 1(3y - 5) = (3y - 5)(2y + 1) \]

### Final Answer
Thus, the factorization of \(6y^2 - 7y - 5\) is:
\[ (3y - 5)(2y + 1) \]

---