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Mathematics

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

2x3 - 9x2 + 7x + 6.

Factorisation

ICSE Sp 2024

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Answer

Substituting x = 2, in the given polynomial, we get :

⇒ 2(2)3 - 9(2)2 + 7(2) + 6

⇒ 2 × 8 - 9 × 4 + 14 + 6

⇒ 16 - 36 + 20

⇒ -20 + 20

⇒ 0.

∴ x - 2 is the factor of the given polynomial.

On dividing 2x3 - 9x2 + 7x + 6 by x - 2, we get :

x−2)2x2−5x−3x−2)2x3−9x2+7x+6‾x−2))+−2x3−+4x2‾x−22x3−4−5x2+7xx−2)2x3−4−+5x2+−10x‾x−2)31x3−2+1−3x+6x−2)31x3−2+11−+3x+−6‾x−2)31x3−2+11+1×\begin{array}{l} \phantom{x - 2)}{\quad 2x^2 - 5x - 3} \ x - 2\overline{\smash{\big)}\quad 2x^3 - 9x^2 + 7x + 6} \ \phantom{x - 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{+}{-}4x^2} \ \phantom{{x - 2}2x^3-4}-5x^2 + 7x \ \phantom{{x - 2)}2x^3-4}\underline{\underset{+}{-}5x^2 \underset{-}{+} 10x} \ \phantom{{x - 2)}31x^3-2+1}-3x + 6 \ \phantom{{x - 2)}31x^3-2+11}\underline{\underset{+}{-}3x \underset{-}{+} 6} \ \phantom{{x - 2)}31x^3-2+11+1}\times \end{array}

∴ 2x3 - 9x2 + 7x + 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 - 9x2 + 7x + 6 = (x - 2)(2x + 1)(x - 3).

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