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Mathematics

Using Remainder and Factor theorem factorise the given polynomial completely.

6x3 + x2 - 4x + 1

Factorisation

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Answer

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 :

x−3x)6x2−5x+1x+1)6x3+x2−4x+1‾x2+4(+−6x3+−6x2‾x2−4x+1)−5x2−4xx2−4x+1))−+5x2−+5x‾x2−4x+1)+24x())x+1x2−4x+1)+24+)+−x+−1‾x2+3x−54)+2x+7×\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).

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