The HCF of `p(x) = 26(6x^(4) - x^(3) - 2x^(2))` and `q(x) = 20 (2x^(6) + 3x^(5) + x^(4))` is

Options

A. `4x^(2)(2x + 1)`
B. `6x^(3) (2x-1)`
C. `6x^(2)(2x+1)`
D. `4x^(2) (2x-1)`

Answer

To find the HCF (Highest Common Factor) of the polynomials \( p(x) = 26(6x^4 - x^3 - 2x^2) \) and \( q(x) = 20(2x^6 + 3x^5 + x^4) \), we will follow these steps:

### Step 1: Factor out the constants and the variable parts from both polynomials.

1. **For \( p(x) \)**:
- The constant is \( 26 \).
- The polynomial part is \( 6x^4 - x^3 - 2x^2 \).
- We can factor out \( 2x^2 \) from the polynomial:
\[ 6x^4 - x^3 - 2x^2 = 2x^2(3x^2 - \frac{1}{2}x - 1) \]
- Therefore,
\[ p(x) = 26 \cdot 2x^2(3x^2 - \frac{1}{2}x - 1) = 52x^2(3x^2 - \frac{1}{2}x - 1) \]

2. **For \( q(x) \)**:
- The constant is \( 20 \).
- The polynomial part is \( 2x^6 + 3x^5 + x^4 \).
- We can factor out \( 2x^4 \):
\[ 2x^6 + 3x^5 + x^4 = 2x^4(x^2 + \frac{3}{2}x + \frac{1}{2}) \]
- Therefore,
\[ q(x) = 20 \cdot 2x^4(x^2 + \frac{3}{2}x + \frac{1}{2}) = 40x^4(x^2 + \frac{3}{2}x + \frac{1}{2}) \]

### Step 2: Identify common factors.

- The constants from \( p(x) \) and \( q(x) \) are \( 52 \) and \( 40 \).
- The HCF of \( 52 \) and \( 40 \) is \( 4 \).
- The variable part from \( p(x) \) is \( x^2 \) and from \( q(x) \) is \( x^4 \). The HCF of \( x^2 \) and \( x^4 \) is \( x^2 \).

### Step 3: Combine the common factors.

- The common factor from the constants is \( 4 \).
- The common factor from the variable parts is \( x^2 \).

Thus, the HCF of \( p(x) \) and \( q(x) \) is:
\[ \text{HCF} = 4x^2 \]

### Final Answer:
The HCF of \( p(x) \) and \( q(x) \) is \( 4x^2 \).

---