What is HCF of `8x^2y^2, 12x^3y^2` and `24x^4 y^3 z^2` ?

Answer

To find the HCF (Highest Common Factor) of the polynomials \(8x^2y^2\), \(12x^3y^2\), and \(24x^4y^3z^2\), we will follow a systematic approach.

### Step 1: Factor each term into its prime factors.

1. **Factor \(8x^2y^2\)**:
- \(8 = 2^3\)
- Thus, \(8x^2y^2 = 2^3 \cdot x^2 \cdot y^2\)

2. **Factor \(12x^3y^2\)**:
- \(12 = 2^2 \cdot 3\)
- Thus, \(12x^3y^2 = 2^2 \cdot 3 \cdot x^3 \cdot y^2\)

3. **Factor \(24x^4y^3z^2\)**:
- \(24 = 2^3 \cdot 3\)
- Thus, \(24x^4y^3z^2 = 2^3 \cdot 3 \cdot x^4 \cdot y^3 \cdot z^2\)

### Step 2: Identify the common factors.

Now we will look for the common factors in each of the factorizations:

- **For the constant factors**:
- \(2^3\) from \(8\) and \(24\), and \(2^2\) from \(12\). The minimum power is \(2^2\).

- **For the variable \(x\)**:
- \(x^2\) from \(8x^2y^2\), \(x^3\) from \(12x^3y^2\), and \(x^4\) from \(24x^4y^3z^2\). The minimum power is \(x^2\).

- **For the variable \(y\)**:
- \(y^2\) from \(8x^2y^2\), \(y^2\) from \(12x^3y^2\), and \(y^3\) from \(24x^4y^3z^2\). The minimum power is \(y^2\).

- **For the variable \(z\)**:
- \(z\) appears only in \(24x^4y^3z^2\) and not in the other two terms, so it will not be included in the HCF.

### Step 3: Combine the common factors.

Now we combine the common factors we identified:

\[ \text{HCF} = 2^2 \cdot x^2 \cdot y^2 \]

### Step 4: Simplify the expression.

Calculating \(2^2\):

\[ \text{HCF} = 4x^2y^2 \]

### Final Answer:

The HCF of \(8x^2y^2\), \(12x^3y^2\), and \(24x^4y^3z^2\) is:

\[ \boxed{4x^2y^2} \]

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