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Mathematics

Using Factor Theorem, show that :

(3x + 2) is a factor of 3x3 + 2x2 - 3x - 2. Hence, factorise the expression 3x3 + 2x2 - 3x - 2 completely.

Factorisation

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Answer

3x + 2 = 0 ⇒ x = −23-\dfrac{2}{3}.

Remainder = The value of 3x3 + 2x2 - 3x - 2 at x = −23-\dfrac{2}{3}.

=3(−23)3+2(−23)2−3(−23)−2=3(−827)+2(49)+2−2=−89+89=−8+89=0.= 3\Big(-\dfrac{2}{3}\Big)^3 + 2\Big(-\dfrac{2}{3}\Big)^2 - 3\Big(-\dfrac{2}{3}\Big) - 2 \\[1em] = 3\Big(-\dfrac{8}{27}\Big) + 2\Big(\dfrac{4}{9}\Big) + 2 - 2 \\[1em] = -\dfrac{8}{9} + \dfrac{8}{9} \\[1em] = \dfrac{-8 + 8}{9} \\[1em] = 0.

Hence, 3x + 2 is a factor of 3x3 + 2x2 - 3x - 2.

Now dividing, 3x3 + 2x2 - 3x - 2 by 3x + 2,

3x+2)x2−13x+2)3x3+2x2−3x−2‾3x+2−3x3+−2x2‾3x+23x3+2x2−3x−23x+23x3+2x2−+3x−+2‾3x+23x3+2x2−3x×\begin{array}{l} \phantom{3x + 2)}{x^2 - 1} \ 3x + 2\overline{\smash{\big)}3x^3 + 2x^2 - 3x - 2} \ \phantom{3x + 2}\underline{\underset{-}{}3x^3 \underset{-}{+}2x^2} \ \phantom{{3x + 2}3x^3+2x^2\enspace}-3x - 2 \ \phantom{{3x + 2}3x^3+2x^2\quad}\underline{\underset{+}{-}3x \underset{+}{-} 2} \ \phantom{{3x + 2}{3x^3+2x^2\enspace}{-3x}}\times \end{array}

we get quotient = x2 - 1.

Factorising, x2 - 1

= (x)2 - (1)2

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

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

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

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