Fractorise:
`25a^(2)-4b^(2) +28bc-49c^(2)`

Answer

To factorize the expression \( 25a^2 - 4b^2 + 28bc - 49c^2 \), we can follow these steps:

### Step 1: Rearrange the Expression
We can rearrange the expression to group the terms involving \( b \) and \( c \):
\[ 25a^2 - (4b^2 - 28bc + 49c^2) \]

### Step 2: Recognize the Quadratic Form
Notice that the expression \( 4b^2 - 28bc + 49c^2 \) can be recognized as a perfect square trinomial. We can rewrite it as:
\[ 4b^2 - 28bc + 49c^2 = (2b - 7c)^2 \]

### Step 3: Substitute Back
Now substituting this back into our expression gives:
\[ 25a^2 - (2b - 7c)^2 \]

### Step 4: Apply the Difference of Squares Formula
We can now apply the difference of squares formula, which states that \( x^2 - y^2 = (x - y)(x + y) \). Here, let \( x = 5a \) and \( y = (2b - 7c) \):
\[ 25a^2 - (2b - 7c)^2 = (5a - (2b - 7c))(5a + (2b - 7c)) \]

### Step 5: Simplify
Now simplify the factors:
\[ (5a - 2b + 7c)(5a + 2b - 7c) \]

### Final Factorized Form
Thus, the factorized form of the expression \( 25a^2 - 4b^2 + 28bc - 49c^2 \) is:
\[ (5a - 2b + 7c)(5a + 2b - 7c) \]

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