Factorise : `6x^(3)-5x^(2)-13x+12`

Options

A. `(x-1)(3x-4)(2x+3)`
B. `(x+1)(3x+4)(2x-3)`
C. `(x-1)(3x-4)(2x-3)`
D. `(x+1)(3x+4)(2x+3)`

Answer

To factorize the polynomial \(6x^3 - 5x^2 - 13x + 12\), we will follow these steps:

### Step 1: Identify the polynomial
Let \(P(x) = 6x^3 - 5x^2 - 13x + 12\).

### Step 2: Use the Hit and Trial Method
We will use the hit and trial method to find a root of the polynomial. We will test \(x = 1\) and \(x = -1\).

#### Testing \(x = 1\):
\[ P(1) = 6(1)^3 - 5(1)^2 - 13(1) + 12 = 6 - 5 - 13 + 12 = 0 \]
Since \(P(1) = 0\), \(x - 1\) is a factor of the polynomial.

### Step 3: Polynomial Division
Now we will divide \(P(x)\) by \(x - 1\).

Using synthetic division or long division, we divide \(6x^3 - 5x^2 - 13x + 12\) by \(x - 1\).

1. Divide the leading term: \(6x^3 \div x = 6x^2\).
2. Multiply \(6x^2\) by \(x - 1\): \(6x^3 - 6x^2\).
3. Subtract:
\[ (6x^3 - 5x^2) - (6x^3 - 6x^2) = x^2 \]
4. Bring down the next term: \(x^2 - 13x\) gives \(x^2 - 13x + 12\).
5. Divide: \(x^2 \div x = x\).
6. Multiply: \(x(x - 1) = x^2 - x\).
7. Subtract:
\[ (x^2 - 13x) - (x^2 - x) = -12x \]
8. Bring down the next term: \(-12x + 12\).
9. Divide: \(-12x \div x = -12\).
10. Multiply: \(-12(x - 1) = -12x + 12\).
11. Subtract:
\[ (-12x + 12) - (-12x + 12) = 0 \]

The result of the division is \(6x^2 + x - 12\).

### Step 4: Factor the Quadratic
Now we need to factor \(6x^2 + x - 12\).

To factor \(6x^2 + x - 12\), we look for two numbers that multiply to \(6 \times -12 = -72\) and add to \(1\). The numbers are \(9\) and \(-8\).

We can rewrite the quadratic:
\[ 6x^2 + 9x - 8x - 12 \]
Now, group the terms:
\[ (6x^2 + 9x) + (-8x - 12) \]
Factor by grouping:
\[ 3x(2x + 3) - 4(2x + 3) \]
Combine:
\[ (3x - 4)(2x + 3) \]

### Step 5: Write the Final Factorization
Now we combine all the factors:
\[ P(x) = (x - 1)(3x - 4)(2x + 3) \]

### Final Answer
The factorization of \(6x^3 - 5x^2 - 13x + 12\) is:
\[ (x - 1)(3x - 4)(2x + 3) \]

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