factorize `2x^(2) - 5xy + 3y^(2)`, and use your result to factorize `2(3a-b)^(2) - 5(3a-b)(2a-b) + 3(2a-b)^(2)`. The factors are

Options

A. `a^(2)b`
B. `(a+b)(a-b)`
C. ab
D. `(2a+b)(a-b)`

Answer

To factorize the expression \(2x^2 - 5xy + 3y^2\), we will follow these steps:

### Step 1: Identify the coefficients
The expression is in the form \(ax^2 + bxy + cy^2\), where:
- \(a = 2\)
- \(b = -5\)
- \(c = 3\)

### Step 2: Find the product and sum
We need to find two numbers that multiply to \(a \cdot c = 2 \cdot 3 = 6\) and add up to \(b = -5\).

### Step 3: Determine the factors
The two numbers that satisfy these conditions are \(-3\) and \(-2\) because:
- \(-3 \cdot -2 = 6\) (product)
- \(-3 + -2 = -5\) (sum)

### Step 4: Rewrite the middle term
We can rewrite the expression by splitting the middle term using these factors:
\[ 2x^2 - 3xy - 2xy + 3y^2 \]

### Step 5: Group the terms
Now we group the terms:
\[ (2x^2 - 3xy) + (-2xy + 3y^2) \]

### Step 6: Factor by grouping
Factor out the common factors in each group:
\[ x(2x - 3y) - y(2x - 3y) \]

### Step 7: Factor out the common binomial
Now we can factor out the common binomial \((2x - 3y)\):
\[ (2x - 3y)(x - y) \]

Thus, the factorized form of \(2x^2 - 5xy + 3y^2\) is:
\[ (2x - 3y)(x - y) \]

### Step 8: Substitute back into the second expression
Now we need to use this result to factorize \(2(3a - b)^2 - 5(3a - b)(2a - b) + 3(2a - b)^2\).

Let:
- \(x = 3a - b\)
- \(y = 2a - b\)

### Step 9: Substitute into the expression
Substituting \(x\) and \(y\) into the factorized form:
\[ (2(3a - b) - 3(2a - b))((3a - b) - (2a - b)) \]

### Step 10: Simplify the factors
Now simplify each factor:
1. For the first factor:
\[ 2(3a - b) - 3(2a - b) = 6a - 2b - 6a + 3b = b \]

2. For the second factor:
\[ (3a - b) - (2a - b) = 3a - b - 2a + b = a \]

### Final Result
Thus, the factors of the expression \(2(3a - b)^2 - 5(3a - b)(2a - b) + 3(2a - b)^2\) are:
\[ b(a) \]