Determine whether (x-5) is a factor of the polynomial $p (x) = 2 x^{3} - 5 x^{2} - 28 x + 15$ .

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Solution

By factor theorem, if p(5)=0, then (x-5) is a factor of p(x). Now,
$p (5) = 2 (5)^{3} - 5 (5)^{2} - 28 (5) + 15$
$= 2 (125) - 5 (25) - 140 + 15$
$= 250 - 125 - 140 + 15 = 0$
$∴ (x - 5)$ is a factor of $p (x) = 2 x^{3} - 5 x^{2} - 28 x + 15$ .

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