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Mathematics

Using remainder and factor theorem, factorize completely, the given polynomial :

2x3 + x2 - 13x + 6

Factorisation

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Answer

Substituting x = 2, in the polynomial 2x3 + x2 - 13x + 6, we get :

⇒ 2(2)3 + (2)2 - 13(2) + 6

⇒ 2(8) + 4 - 26 + 6

⇒ 16 + 4 - 20

⇒ 20 - 20

⇒ 0.

∴ x - 2 is the factor of the polynomial 2x3 + x2 - 13x + 6.

Dividing 2x3 + x2 - 13x + 6 by (x - 2), we get,

x−2)2x2+5x−3x−2)2x3+x2−13x+6‾x−2))+−2x3−+4x2‾x−223x3+x25x2−13xx−2)2x3−2+−5x2−+10x‾x−2)23x3−2x2(3)−3x−6x−2)23x3−2x2(31)−+3x−+6‾x−2)23x3−2x2(31)−2x×\begin{array}{l} \phantom{x - 2)}{\quad 2x^2 +5x - 3} \ x - 2\overline{\smash{\big)}\quad 2x^3 + x^2 - 13x + 6} \ \phantom{x - 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{+}{-}4x^2} \ \phantom{{x - 2}23x^3 + x^2}5x^2 - 13x \ \phantom{{x - 2)}2x^3-2}\underline{\underset{-}{+}5x^2 \underset{+}{-} 10x} \ \phantom{{x - 2)}{23x^3-2x^{2}(3)}}-3x - 6 \ \phantom{{x - 2)}{23x^3-2x^{2}(31)}}\underline{\underset{+}{-}3x \underset{+}{-} 6} \ \phantom{{x - 2)}{23x^3-2x^{2}(31)}{-2x}}\times \end{array}

∴ 2x3 + x2 - 13x + 6 = (x - 2)(2x2 + 5x - 3)

= (x - 2)[2x2 + 6x - x - 3]

= (x - 2)[2x(x + 3) - 1(x + 3)]

= (x - 2)(2x - 1)(x + 3).

Hence, 2x3 + x2 - 13x + 6 = (x - 2)(2x - 1)(x + 3).

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