The polynomial 3x³ + 8x² - 15x + k has (x - 1) as a factor. Find the value of k. Hence factorize the resulting polynomial completely.

⇒ x - 1 = 0

⇒ x = 1.

Given, (x - 1) is a factor of 3x³ + 8x² - 15x + k.

Thus, on substituting x = 1 in 3x³ + 8x² - 15x + k, the remainder will be zero.

⇒ 3.(1)³ + 8.(1)² - 15(1) + k = 0

⇒ 3.1 + 8.1 - 15 + k = 0

⇒ 3 + 8 - 15 + k = 0

⇒ 11 - 15 + k = 0

⇒ k - 4 = 0

⇒ k = 4.

Polynomial = 3x³ + 8x² - 15x + 4

On dividing (3x³ + 8x² - 15x + 4) by (x - 1), we get :

$\begin{array}{l} \phantom{x - 1)}{\quad 3x^2 + 11x - 4} \ x - 1\overline{\smash{\big)}\quad 3x^3 + 8x^2 - 15x + 4} \ \phantom{x - 1)}\phantom{)}\underline{\underset{-}{+}3x^3 \underset{+}{-}3x^2} \ \phantom{{x - 1}31x^3-2}11x^2 - 15x \ \phantom{{x - 1)}x^3-2}\underline{\underset{-}{+}11x^2 \underset{+}{-} 11x} \ \phantom{{x - 1)}31x^3-2+1}-4x + 4 \ \phantom{{x - 1)}31x^3-2+11}\underline{\underset{+}{-}4x \underset{-}{+} 4} \ \phantom{{x - 1)}31x^3-2+11+1}\times \end{array}$

⇒ 3x³ + 8x² - 15x + 4 = (x - 1)(3x² + 11x - 4)

= (x - 1)[3x² + 12x - x - 4]

= (x - 1)[3x(x + 4) - 1(x + 4)]

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

Hence, 3x³ + 8x² - 15x + 4 = (x - 1)(3x - 1)(x + 4).