The polynomial px³ + 4x² - 3x + q is completely divisible by x² - 1; find the values of p and q. Also for these values of p and q factorize the given polynomial completely.

x² - 1 is a factor of px³ + 4x² - 3x + q.

∴ (x - 1) and (x + 1) are factors of px³ + 4x² - 3x + q.

Hence, substituting x = 1, -1 remainder = 0..

⇒ p(1)³ + 4(1)² - 3(1) + q = 0

⇒ p + 4 - 3 + q = 0

⇒ p + q = -1

⇒ p = -1 - q …….(i)

p(-1)³ + 4(-1)² - 3(-1) + q = 0

⇒ -p + 4 + 3 + q = 0

⇒ p = 7 + q …….(ii)

From (i) and (ii) we get,

⇒ -1 - q = 7 + q

⇒ 2q = -1 - 7

⇒ 2q = -8

⇒ q = -4.

Substituting q = -4 in (i) we get,

⇒ p = -1 - (-4) = -1 + 4 = 3.

Substituting p = 3 and q = -4 in px³ + 4x² - 3x + q,

= 3x³ + 4x² - 3x - 4.

On dividing, 3x³ + 4x² - 3x - 4 by x - 1,

$\begin{array}{l} \phantom{x - 1)}{3x^2 + 7x + 4} \ x - 1\overline{\smash{\big)}3x^3 + 4x^2 - 3x - 4} \ \phantom{x - 1}\underline{\underset{-}{}3x^3 \underset{+}{-} 3x^2} \ \phantom{{x - 1}3x^3+3}7x^2 - 3x \ \phantom{{x - 1}3x^3+}\underline{\underset{-}{}7x^2 \underset{+}{-} 7x} \ \phantom{{x - 1}{3x^3+3x^2+}{+2}}4x - 4 \ \phantom{{x - 1}{3x^3+3x^2+}\enspace \space}\underline{\underset{-}{}4x \underset{+}{-} 4} \ \phantom{{x - 1}{3x^3+3x^2+}{+2-}}\times \end{array}$

we get, quotient = 3x² + 7x + 4

Factorising 3x² + 7x + 4,

= 3x² + 3x + 4x + 4

= 3x(x + 1) + 4(x + 1)

= (3x + 4)(x + 1).

Hence, p = 3, q = -4 and 3x³ + 4x² - 3x - 4 = (x - 1)(x + 1)(3x + 4).