The HCF of `p(x) = 24 (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 of the polynomials \( p(x) = 24(6x^4 - x^3 - 2x^2) \) and \( q(x) = 20(2x^6 + 3x^5 + x^4) \), we will follow these steps:

### Step 1: Factorize \( p(x) \)
1. Start with \( p(x) = 24(6x^4 - x^3 - 2x^2) \).
2. Factor out \( 24x^2 \):
\[ p(x) = 24x^2(6x^2 - x - 2) \]

3. Now, we need to factor \( 6x^2 - x - 2 \). We can do this by splitting the middle term:
- We look for two numbers that multiply to \( 6 \times (-2) = -12 \) and add to \( -1 \). The numbers are \( -4 \) and \( 3 \).
- Rewrite \( -x \) as \( -4x + 3x \):
\[ 6x^2 - 4x + 3x - 2 \]
- Group the terms:
\[ = 2x(3x - 2) + 1(3x - 2) \]
- Factor out \( (3x - 2) \):
\[ = (3x - 2)(2x + 1) \]

4. Therefore, the complete factorization of \( p(x) \) is:
\[ p(x) = 24x^2(3x - 2)(2x + 1) \]

### Step 2: Factorize \( q(x) \)
1. Start with \( q(x) = 20(2x^6 + 3x^5 + x^4) \).
2. Factor out \( 20x^4 \):
\[ q(x) = 20x^4(2x^2 + 3x + 1) \]

3. Now, we need to factor \( 2x^2 + 3x + 1 \). We look for two numbers that multiply to \( 2 \times 1 = 2 \) and add to \( 3 \). The numbers are \( 2 \) and \( 1 \).
- Rewrite \( 3x \) as \( 2x + 1x \):
\[ 2x^2 + 2x + 1x + 1 \]
- Group the terms:
\[ = 2x(x + 1) + 1(x + 1) \]
- Factor out \( (x + 1) \):
\[ = (x + 1)(2x + 1) \]

4. Therefore, the complete factorization of \( q(x) \) is:
\[ q(x) = 20x^4(2x + 1)(x + 1) \]

### Step 3: Find the HCF
1. The factorizations are:
- \( p(x) = 24x^2(3x - 2)(2x + 1) \)
- \( q(x) = 20x^4(2x + 1)(x + 1) \)

2. Identify the common factors:
- The common numerical factor is \( \text{HCF}(24, 20) = 4 \).
- The common variable factor is \( x^2 \) (since \( x^2 \) is the lowest power of \( x \) present in both).
- The common polynomial factor is \( (2x + 1) \).

3. Therefore, the HCF of \( p(x) \) and \( q(x) \) is:
\[ \text{HCF} = 4x^2(2x + 1) \]

### Final Answer
Thus, the HCF of \( p(x) \) and \( q(x) \) is:
\[ \text{HCF} = 4x^2(2x + 1) \]