Factorise:
`m ^(2) - 4 mn + 4n^(2)`

Answer

To factorise the expression \( m^2 - 4mn + 4n^2 \), we can follow these steps:

### Step 1: Identify the structure of the expression
The expression \( m^2 - 4mn + 4n^2 \) resembles the standard form of a perfect square trinomial, which is \( a^2 - 2ab + b^2 = (a - b)^2 \).

### Step 2: Rewrite the expression
We can rewrite \( 4n^2 \) as \( (2n)^2 \). Thus, the expression becomes:
\[ m^2 - 4mn + (2n)^2 \]

### Step 3: Identify \( a \) and \( b \)
In our case, we can identify:
- \( a = m \)
- \( b = 2n \)

### Step 4: Apply the perfect square formula
Now we can apply the perfect square formula:
\[ a^2 - 2ab + b^2 = (a - b)^2 \]
Substituting \( a \) and \( b \):
\[ m^2 - 2(m)(2n) + (2n)^2 = (m - 2n)^2 \]

### Step 5: Write the final factorised form
Thus, the factorised form of the expression \( m^2 - 4mn + 4n^2 \) is:
\[ (m - 2n)^2 \]

### Summary of the factorisation
The final answer is:
\[ m^2 - 4mn + 4n^2 = (m - 2n)^2 \]

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