What is the HCF of (x3 - 4x2 + x + 6) and (x3 - 7x - 6)?

Given:

Polynomial 1: (x³ - 4x² + x + 6)

Polynomial 2: (x³ - 7x - 6)

Formula Used:

HCF of polynomials by factorization.

Calculation:

Factorize Polynomial 1: (x³ - 4x² + x + 6)

By trial and error, if x = -1,

⇒ (-1)³ - 4(-1)² + (-1) + 6 = -1 - 4 - 1 + 6 = 0

So, (x + 1) is a factor.

Dividing (x³ - 4x² + x + 6) by (x + 1) gives (x² - 5x + 6).

Factorizing (x² - 5x + 6):

⇒ x² - 2x - 3x + 6

⇒ x(x - 2) - 3(x - 2)

⇒ (x - 2)(x - 3)

So, Polynomial 1 = (x + 1)(x - 2)(x - 3)

Factorize Polynomial 2: (x³ - 7x - 6)

By trial and error, if x = -1,

⇒ (-1)³ - 7(-1) - 6 = -1 + 7 - 6 = 0

So, (x + 1) is a factor.

Dividing (x³ - 7x - 6) by (x + 1) gives (x² - x - 6).

Factorizing (x² - x - 6):

⇒ x² - 3x + 2x - 6

⇒ x(x - 3) + 2(x - 3)

⇒ (x - 3)(x + 2)

So, Polynomial 2 = (x + 1)(x - 3)(x + 2)

Common factors of Polynomial 1 and Polynomial 2 are (x + 1) and (x - 3).

HCF = (x + 1)(x - 3)

⇒ x² - 3x + x - 3

⇒ x² - 2x - 3

The correct answer is Option (3).