If $x^{2} + 5 x - 2 k$ is exactly divisible by $(x - 1)$ then the value of k is

1

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Solution

The correct option is C $3$
Given that,
when $(x - 1)$ is factor of $x^{2}$ + $5 x - 2 k$ .
$(1)^{2} + 5 \times$ $1 - 2 k = 0$
$1 + 5 - 2 k = 0$
$6 - 2 k = 0$
$2 k = 6$ $\Rightarrow$ $k = 3$
Putting $x = 1$ . the remainder will be equal to zero.

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