We are given the expression to factorize:

\[
6(a-b)^2 - 11(a-b) + 3
\]

To factor this, we'll treat \( (a-b) \) as a single variable, say \( x \). So the expression becomes:

\[
6x^2 - 11x + 3
\]

Now, we factor this quadratic expression using the following steps:

### Step 1: Multiply the coefficient of \( x^2 \) (which is 6) by the constant term (which is 3):

\[
6 \times 3 = 18
\]

### Step 2: Find two numbers that multiply to 18 and add to -11. These numbers are -9 and -2, since:

\[
-9 \times -2 = 18 \quad \text{and} \quad -9 + -2 = -11
\]

### Step 3: Rewrite the middle term using -9 and -2:

\[
6x^2 - 9x - 2x + 3
\]

### Step 4: Factor by grouping:

\[
3x(2x - 3) - 1(2x - 3)
\]

### Step 5: Factor out the common term \( (2x - 3) \):

\[
(3x - 1)(2x - 3)
\]

### Step 6: Substitute \( x = (a-b) \) back into the expression:

\[
(3(a-b) - 1)(2(a-b) - 3)
\]

Therefore, the factorized form of the given expression is:

\[
(3(a-b) - 1)(2(a-b) - 3)
\]

Would you like more details or have any questions?

Here are 5 related questions for practice:
1. Factorize \( 8(x-2)^2 - 22(x-2) + 5 \).
2. Simplify \( (x-y)(3x + 2y) - (2x + y)(x - y) \).
3. Solve \( 3x^2 - 11x + 6 = 0 \) using the factorization method.
4. Factorize \( 4p^2 - 12p + 9 \).
5. Expand and simplify \( (2a - 3b)(4a + b) \).

**Tip:** When factorizing quadratics, always check for a common factor before attempting other methods.

Learn how to factorize the expression 6(a-b)² - 11(a-b) + 3 using quadratic factoring methods. This detailed step-by-step solution explores key algebraic concepts such as factoring by grouping, quadratic expressions, and substitution. Suitable for students in Grades 9-11.

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