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ICSE Class 10 Mathematics Question 20 of 21

Solved 2023 Question Paper ICSE Class 10 Mathematics — Question 19

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19
Question

Question 10(i)

Factorize completely using factor theorem : 2x3 - x2 - 13x - 6.

Answer

Substituting x = -2 in 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 a 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+2x3−2−5x2−13xx+2)x3−2−+5x2−+10x‾x+2)x3−2x2(3)−3x−6x+2)x3−2x2(31)−+3x−+6‾x+2)x3−2x2(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}x^3-2}-5x^2 - 13x \\ \phantom{{x + 2)}x^3-2}\underline{\underset{+}{-}5x^2 \underset{+}{-} 10x} \\ \phantom{{x + 2)}{x^3-2x^{2}(3)}}-3x - 6 \\ \phantom{{x + 2)}{x^3-2x^{2}(31)}}\underline{\underset{+}{-}3x \underset{+}{-} 6} \\ \phantom{{x + 2)}{x^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).