Find the value of $k$ , if $2 x - 3$ is a factor of $2 x^{3} - k x^{2} + x + 12$ .

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Solution

$(2 x - 3)$ is a factor of $p (x) = 2 x^{3} - 9 x^{2} + x + K$ ,
If $(2 x - 3) = 0, x = \frac{3}{2}$
If $(2 x - 3)$ is a factor of $p (x)$ then $p (\frac{3}{2}) = 0$
$p (x) = 2 x^{3} - 9 x^{2} + x + K$ ,
$\Rightarrow p (\frac{3}{2}) = 2 (\frac{3}{2}) - 9 (\frac{3}{2}) + \frac{3}{2} + K = 0$
$\Rightarrow 2 (\frac{27}{8}) - 9 (\frac{9}{4}) + \frac{3}{2} + K = 0$
$\Rightarrow (\frac{27}{4} - \frac{81}{4} + \frac{3}{2} + K = 0) \times 4$
$27 - 81 + 6 + 4 K = 0$
$- 48 + 4 K = 0$
$4 K = 48$
So $K = 12$

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Find the value of $k$ if $(2 x - 3)$ is a factor of $2 x^{3} - 9 x^{2} + x + k .$

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Find the value of k, if 2x−3 is a factor of 2x3−kx2+x+12.

(2x−3) is a factor of p(x)=2x3−9x2+x+K,If (2x−3)=0,x=32If (2x−3) is a factor of p(x) then p(32)=0p(x)=2x3−9x2+x+K,⇒p(32)=2(32)−9(32)+32+K=0⇒2(278)−9(94)+32+K=0⇒(274−814+32+K=0)×427−81+6+4K=0−48+4K=04K=48So K=12

Find the value of $k$ , if $2 x - 3$ is a factor of $2 x^{3} - k x^{2} + x + 12$ .