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Mathematics

Use factor theorem to factorise the following polynomials completely :

6x3 + 17x2 + 4x - 12

Factorisation

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Answer

Given,

f(x) = 6x3 + 17x2 + 4x - 12

Substituting x = -2 in f(x), we get :

f(-2) = 6(-2)3 + 17(-2)2 + 4(-2) - 12

= 6 × -8 + 17 × 4 - 8 - 12

= -48 + 68 - 8 - 12

= -68 + 68

= 0.

Since, f(-2) = 0, hence (x + 2) is factor of f(x).

Dividing f(x) by (x + 2), we get :

x−3x)6x2+5x−6x+2)6x3+17x2+4x−12‾x2+4))+−6x3+−12x2‾x2+3x−5=d)5x2+4x−12x2+3x−54=)+−5x2+−10x‾x2+3x−54)+2x+3−6x−12x2+3x−54=zccdvz)−+6x−+12‾x2+3x−54)+2x+3+dc×\begin{array}{l} \phantom{x - 3x )}{\quad 6x^2 + 5x - 6} \ x + 2\overline{\smash{\big)}\quad 6x^3 + 17x^2 + 4x - 12} \ \phantom{x^2 + 4)}\phantom{)}\underline{\underset{-}{+}6x^3 \underset{-}{+} 12x^2 } \ \phantom{{x^2 + 3x - 5 =d)}} 5x^2 + 4x - 12 \ \phantom{{x^2 + 3x - 54 =)}}\underline{\underset{-}{+}5x^2 \underset{-}{+}10x } \ \phantom{{x^2 + 3x - 54)} + 2x + 3 }-6x - 12\ \phantom{{x^2 + 3x - 54 =zccdvz)}}\underline {\underset{+}{-}6x \underset{+}{-}12 } \ \phantom{{x^2 + 3x - 54)} + 2x + 3 + dc}\times\ \end{array}

We get 6x2 + 5x - 6 as the quotient and remainder = 0.

∴ 6x3 + 17x2 + 4x - 12 = (x + 2)(6x2 + 5x - 6)

= (x + 2)(6x2 + 9x - 4x - 6)

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

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

Hence,6x3 + 17x2 + 4x - 12 = (x + 2)(2x + 3)(3x - 2).

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