If 2x³ - 3x² - 3x + 2 = (2x - 1)(x² + ax + b)

(a) using Remainder and Factor theorem, find the value of ‘a’ and ‘b’.

(b) hence, factorise the polynomial 2x³ - 3x² - 3x + 2 completely.

(a) Given,

2x³ - 3x² - 3x + 2 = (2x - 1)(x² + ax + b)

Therefore, (2x - 1) is factor of 2x³ - 3x² - 3x + 2.

Factorizing,

$\begin{array}{l} \phantom{x - 3x)}{\quad x^2 - x - 2} \ 2x - 1\overline{\smash{\big)}\quad 2x^3 - 3x^2 - 3x + 2 } \ \phantom{x^2 + 4}\phantom(\underline{\underset{-}{+}2x^3 \underset{+}{-}x^2} \ \phantom{x^2 + 3x - 7)} - 2x^2 - 3x \ \phantom{x^2 + 3x - 5))}\underline{\underset{+}{-}2x^2 \underset{-}{+}x} \ \phantom{x^2 + 3x - 5) + 24)}-4x + 2 \ \phantom{{x^2 + 3x - 5) + 24 +}}\underline{\underset{+}{-}4x \underset{-}{+}2} \ \phantom{{x^2 + 3x - 54)} + 2x + 7} \times\ \end{array}$

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

Comparing, x² + ax + b with x² - x - 2, we get:

a = -1 and b = -2.

Hence, a = -1 and b = -2.

(b) From part (a),

⇒ 2x³ - 3x² - 3x + 2 = (2x - 1)(x² - x - 2)

= (2x - 1)(x² - 2x + x - 2)

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

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

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