Factorise:
`14 x ^(3) + 21x ^(4) y - 28 x ^(2) y ^(2)`

Answer

To factorise the expression \( 14x^3 + 21x^4y - 28x^2y^2 \), we will follow these steps:

### Step 1: Identify the common factors
First, we need to identify the common factors in all the terms of the expression. The coefficients are 14, 21, and -28. The greatest common divisor (GCD) of these numbers is 7.

Next, we look at the variable part. The terms are \( x^3 \), \( x^4y \), and \( x^2y^2 \). The lowest power of \( x \) is \( x^2 \), and the lowest power of \( y \) is \( y^0 \) (which means \( y \) is not present in the first term). Thus, we can take \( x^2 \) as a common factor.

### Step 2: Factor out the common factor
Now we can factor out \( 7x^2 \) from the expression:

\[ 14x^3 + 21x^4y - 28x^2y^2 = 7x^2(2x) + 7x^2(3x^2y) - 7x^2(4y^2) \]

### Step 3: Rewrite the expression
After factoring out \( 7x^2 \), we can rewrite the expression inside the parentheses:

\[ = 7x^2(2x + 3x^2y - 4y^2) \]

### Step 4: Check for further factorization
Now we need to check if the expression \( 2x + 3x^2y - 4y^2 \) can be factored further. In this case, it does not seem to have any common factors or be factorable into simpler binomials.

Thus, the final factorized form of the expression is:

\[ \boxed{7x^2(2x + 3x^2y - 4y^2)} \]