For the remainder of the division of $x^{3} - 2 x^{2} + 3 k x + 18$ by $x - 6$ to be equal to zero, k must be equal to

0

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1

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5

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\frac{27}{2}

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\frac{- 27}{2}

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Solution

The correct option is E $\frac{- 27}{2}$
Remainder theorem: remainder when polynomial $P (x)$ is divided by $(x - a)$ is $P (a)$

Therefore remainder when $P (x) = x^{3} - 2 x^{2} + 3 k x + 18$ is divided by $x - 6$ is
$= P (6) = 6^{3} - 2 (6^{2}) + 2 k (6) + 18 = 0$ (remainder is given zero here )
$\Rightarrow 36 - 12 + 2 k + 3 = 0$
$\Rightarrow k = - \frac{27}{2}$

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For the remainder of the division of x3−2x2+3kx+18 by x−6 to be equal to zero, k must be equal to

The correct option is E −272Remainder theorem: remainder when polynomial P(x) is divided by (x−a) is P(a)Therefore remainder when P(x)=x3−2x2+3kx+18 is divided by x−6 is=P(6)=63−2(62)+2k(6)+18=0 (remainder is given zero here )⇒36−12+2k+3=0⇒k=−272

For the remainder of the division of $x^{3} - 2 x^{2} + 3 k x + 18$ by $x - 6$ to be equal to zero, k must be equal to