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