Factorise: `6x^2+17x+12`

Answer

To factorize the polynomial \(6x^2 + 17x + 12\), we can follow these steps:

### Step 1: Identify the coefficients
The given polynomial is \(6x^2 + 17x + 12\). Here, the coefficients are:
- \(a = 6\) (coefficient of \(x^2\))
- \(b = 17\) (coefficient of \(x\))
- \(c = 12\) (constant term)

### Step 2: Multiply \(a\) and \(c\)
We need to multiply \(a\) and \(c\):
\[ a \cdot c = 6 \cdot 12 = 72 \]

### Step 3: Find two numbers that multiply to \(72\) and add to \(17\)
Next, we need to find two numbers that multiply to \(72\) and add up to \(17\). The numbers \(8\) and \(9\) satisfy this condition:
\[ 8 \cdot 9 = 72 \quad \text{and} \quad 8 + 9 = 17 \]

### Step 4: Rewrite the middle term
We can rewrite the polynomial by splitting the middle term \(17x\) into \(8x + 9x\):
\[ 6x^2 + 8x + 9x + 12 \]

### Step 5: Group the terms
Now, we group the terms:
\[ (6x^2 + 8x) + (9x + 12) \]

### Step 6: Factor out the common factors
From the first group \(6x^2 + 8x\), we can factor out \(2x\):
\[ 2x(3x + 4) \]
From the second group \(9x + 12\), we can factor out \(3\):
\[ 3(3x + 4) \]

### Step 7: Combine the factored terms
Now we have:
\[ 2x(3x + 4) + 3(3x + 4) \]
We can see that \((3x + 4)\) is common in both terms, so we factor it out:
\[ (3x + 4)(2x + 3) \]

### Final Result
Thus, the factorization of the polynomial \(6x^2 + 17x + 12\) is:
\[ \boxed{(3x + 4)(2x + 3)} \]