Using Remainder and Factor theorem factorise the given polynomial completely.
6x3 + x2 - 4x + 1
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Given,
f(x) = 6x3 + x2 - 4x + 1
Substituting x = -1 in f(x), we get :
f(-1) = 6(-1)3 + (-1)2 - 4(-1) + 1
= 6 × -1 + 1 + 4 + 1
= -6 + 1 + 4 + 1
= -6 + 6
= 0.
Since, f(-1) = 0, hence (x + 1) is factor of f(x).
Dividing f(x) by x + 1, we get :
$$\begin{array}{l} \phantom{x - 3x)}{\quad 6x^2 - 5x + 1} \\ x + 1\overline{\smash{\big)}\quad 6x^3 + x^2 - 4x + 1 } \\ \phantom{x^2 + 4}\phantom(\underline{\underset{-}{+}6x^3 \underset{-}{+}6x^2} \\ \phantom{x^2 - 4x + 1)} - 5x^2 - 4x \\ \phantom{x^2 - 4x + 1))}\underline{\underset{+}{-}5x^2 \underset{+}{-}5x} \\ \phantom{x^2 - 4x + 1) + 24x ())}x + 1 \\ \phantom{{x^2 - 4x + 1) + 24 +)}}\underline{\underset{-}{+}x \underset{-}{+}1} \\ \phantom{{x^2 + 3x - 54)} + 2x + 7} \times\ \end{array}$$
∴ 6x3 + x2 - 4x + 1 = (x + 1)(6x2 - 5x + 1)
= (x + 1)(6x2 - 5x + 1)
= (x + 1)(6x2 - 2x - 3x + 1)
= (x + 1)[2x(3x - 1) - 1(3x - 1)]
= (x + 1)(2x - 1)(3x - 1).
Hence, 6x3 + x2 - 4x + 1 = (x + 1)(2x - 1)(3x - 1).