Is $(2 x - 3)$ a factor of $p (x) = 2 x^{3} - 9 x^{2} + x + 10$ ?

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Solution

If $2 x - 3$ is the factor of $p (x)$ ,

then $2 x - 3 = 0$ $⟹ x = \frac{3}{2}$ , must be the root of $p (x)$ .

Substitute this in $p (x)$ , we get,

$p (\frac{3}{2}) =$ $2 (\frac{3}{2})^{3} - 9 (\frac{3}{2})^{2} + \frac{3}{2} + 10$

$= \frac{27}{4} - \frac{81}{4} + \frac{46}{4} = - 2$

So, $p (\frac{3}{2}) \neq 0$

Hence, $(2 x - 3)$ is not the factor of $p (x)$ .

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Similar questions

2 x^{3} - 9 x^{2} + x = 12

Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = 2 x^{3} + 9 x^{2} - 11 x - 30, g (x) = x + 5$

(2 x - 3)

2 x^{3} - 9 x^{2} + x + k

k

69 + 11 x - x^{2} + x^{3}

x + 2 x^{3} - 9 x^{2} + 12

Question 16
Show that:
(i) x + 3 is a factor of $69 + 11 x - x^{2} + x^{3}$ .

(ii) 2x – 3 is a factor of $x + 2 x^{3} - 9 x^{2} + 12$ .

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