Factorise by taking out the common factors :
`2x(a - b) + 3y(5a-5b) + 4z(2b - 2a)`

Answer

To factorise the expression \(2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a)\), we will follow these steps:

### Step 1: Rewrite the expression
First, let's rewrite the expression clearly:
\[ 2x(a - b) + 3y(5a - 5b) + 4z(2b - 2a) \]

### Step 2: Factor out common factors from each term
In the second term \(3y(5a - 5b)\), we can factor out \(5\):
\[ 3y(5a - 5b) = 3y \cdot 5(a - b) = 15y(a - b) \]

In the third term \(4z(2b - 2a)\), we can factor out \(2\) and also rearrange the terms:
\[ 4z(2b - 2a) = 4z \cdot 2(b - a) = -8z(a - b) \]
(Note: \(b - a\) can be rewritten as \(-(a - b)\))

### Step 3: Rewrite the expression with factored terms
Now, substituting these factored forms back into the expression, we have:
\[ 2x(a - b) + 15y(a - b) - 8z(a - b) \]

### Step 4: Factor out the common factor \((a - b)\)
Now we can see that \((a - b)\) is a common factor in all three terms:
\[ (a - b)(2x + 15y - 8z) \]

### Final Answer
Thus, the factorised form of the expression is:
\[ (a - b)(2x + 15y - 8z) \]

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