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.
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(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).