If $(x - 2)$ is a factor of $2 x^{3} - x^{2} - p x - 2$ ,
(i) find the value of $p$
(ii) with the value of $p$ , factorise the above expression completely.

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Solution

Let
$f (x) = 2 x^{3} - x^{2} - p x - 2$

(i) Given
$(x - 2)$ is a factor of $f (x)$
$∴$ By factor theorem, $f (2) = 0$
$\Rightarrow 2 (2)^{3} - (2)^{2} - p (2) - 2 = 0 \Rightarrow 16 - 4 - 2 p - 2 = 0 \Rightarrow 10 = 2 p \Rightarrow p = 5$

(ii) We have
$f (x) = 2 x^{3} - x^{2} - 5 x - 2$
To find the other factor, we divide $f (x)$ by $(x - 2)$ by long division method
$∴ f (x) = (x - 2) (2 x^{2} + 3 x + 1) = (x - 2) (2 x^{2} + 2 x + x + 1) = (x - 2) {2 x (x + 1) + (x + 1)} = (x - 2) (x + 1) (2 x + 1)$

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If (x−2) is a factor of 2x3−x2−px−2,(i) find the value of p(ii) with the value of p, factorise the above expression completely.

If $(x - 2)$ is a factor of $2 x^{3} - x^{2} - p x - 2$ ,
(i) find the value of $p$
(ii) with the value of $p$ , factorise the above expression completely.