Examine whether $x - 2$ is a factor of $2 x^{3} - 3 x^{2} - 5 x + 6$ or not using factor theorem.

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Solution

Let $p (x) = 2 x^{3} - 3 x^{2} - 5 x + 6$ .

Put divisor equal to 0,

$x - 2 = 0$

$x = 2$

Put the value of $x$ in $p (x)$ ,

$p (2) = 2 {(2)}^{3} - 3 {(2)}^{2} - 5 (2) + 6$

$= 16 - 12 - 10 + 6$

$= 0$

Since the remainder is 0, so, $x - 2$ is the factor of $2 x^{3} - 3 x^{2} - 5 x + 6$ .

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