Factor this polynomial completely.

[tex]\[ 12x^2 + x - 6 \][/tex]

A. \((3x - 2)(4x + 3)\)

B. \((12x - 2)(x + 3)\)

C. \((4x - 2)(2x + 3)\)

D. [tex]\((12x - 3)(x + 2)\)[/tex]



Answer :

To factor the polynomial \(12x^2 + x - 6\) completely, let's follow a systematic approach.

### Step 1: Identification of the polynomial
We have the polynomial given as:
[tex]\[ 12x^2 + x - 6 \][/tex]

### Step 2: Finding the factors
We need to express this polynomial as a product of two binomials of the form \((ax + b)(cx + d)\).

### Step 3: Verifying the given choices
We need to determine which among the given options will result in the original polynomial when multiplied out.

Option A: \((3x - 2)(4x + 3)\)
[tex]\[ \begin{aligned} (3x - 2)(4x + 3) & = 3x \cdot 4x + 3x \cdot 3 - 2 \cdot 4x - 2 \cdot 3 \\ & = 12x^2 + 9x - 8x - 6 \\ & = 12x^2 + x - 6 \end{aligned} \][/tex]
This indeed matches our original polynomial.

Option B: \((12x - 2)(x + 3)\)
[tex]\[ \begin{aligned} (12x - 2)(x + 3) & = 12x \cdot x + 12x \cdot 3 - 2 \cdot x - 2 \cdot 3 \\ & = 12x^2 + 36x - 2x - 6 \\ & = 12x^2 + 34x - 6 \end{aligned} \][/tex]
This does not match our original polynomial.

Option C: \((4x - 2)(2x + 3)\)
[tex]\[ \begin{aligned} (4x - 2)(2x + 3) & = 4x \cdot 2x + 4x \cdot 3 - 2 \cdot 2x - 2 \cdot 3 \\ & = 8x^2 + 12x - 4x - 6 \\ & = 8x^2 + 8x - 6 \end{aligned} \][/tex]
This does not match our original polynomial.

Option D: \((12x - 3)(x + 2)\)
[tex]\[ \begin{aligned} (12x - 3)(x + 2) & = 12x \cdot x + 12x \cdot 2 - 3 \cdot x - 3 \cdot 2 \\ & = 12x^2 + 24x - 3x - 6 \\ & = 12x^2 + 21x - 6 \end{aligned} \][/tex]
This also does not match our original polynomial.

### Conclusion
Of the given options, only Option A:
[tex]\[ (3x - 2)(4x + 3) \][/tex]
when multiplied, gives us the original polynomial \(12x^2 + x - 6\).

Therefore, the completely factored form of the polynomial \(12x^2 + x - 6\) is:
[tex]\[ (3x - 2)(4x + 3) \][/tex]
Correspondingly, the correct answer is choice:
[tex]\[ \boxed{A} \][/tex]