If (x - 2) is a factor of 2x³ - x² - px - 2, then

(i) Find the value of p.

(ii) with this value of p, factorise the above expression completely.

(i) f(x) = 2x³ - x² - px - 2

If, (x - 2) is a factor of f(x), then f(2) = 0

$\therefore 2(2)^3 - (2)^2 - 2p - 2 = 0 \\[0.5em] \Rightarrow 16 - 4 - 2 - 2p = 0 \\[0.5em] \Rightarrow 10 - 2p = 0 \\[0.5em] \Rightarrow 2p = 10 \\[0.5em] p = 5.$

Hence, the value of p is 5.

(ii) Putting value of p = 5 in f(x),

f(x) = 2x³ - x² - 5x - 2

Since, (x - 2) is a factor of f(x), dividing f(x) by (x - 2),

$\begin{array}{l} \phantom{x - 2)}{2x^2 + 3x + 1} \ x - 2\overline{\smash{\big)}2x^3 - x^2 - 5x - 2} \ \phantom{x - 2}\underline{\underset{-}{ }2x^3 \underset{+}{-} 4x^2} \ \phantom{{x - 2}2x^3+4}3x^2 - 5x \ \phantom{{x - 2}2x^3+}\underline{\underset{-}{}3x^2 \underset{+}{-} 6x} \ \phantom{{x - 2}{2x^3+}{+3x^2+5}}x - 2 \ \phantom{{x - 2}{2x^3+}{+3x^2+5}}\underline{\underset{-}{ }x \underset{+}{-} 2} \ \phantom{{x - 2}{2x^3+}{+2x^2-}{-4x}}\times \end{array}$

we get, 2x² + 3x + 1 as quotient and remainder = 0.

$\therefore 2x^3 - x^2 - 5x - 2 = (x - 2)(2x^2 + 3x + 1) \\[0.5em] = (x - 2)(2x^2 + 2x + x + 1) \\[0.5em] = (x - 2)(2x(x + 1) + 1(x + 1)) \\[0.5em] = (x - 2)(2x + 1)(x + 1)$

Hence, 2x³ - x² - 5x - 2 = (x - 2)(2x + 1)(x + 1).