MCQ
If $x^3 + 6x^2 + 4x +k$ is exactly divisible by $x + 2$ then $k =$
- A$-8$
- ✓$-7$
- C$-6$
- D$-10$
$-8$
$f(x) = x^3 + 6x^2 + 4x + k$
$f(-2) = 0$
$\therefore (-2)^3 + 6(-2)^2 + 4(-2) + k = 0$
$\therefore -8 + 6(4) + (-8) + k = 0$
$24 - 16 + k = 0$
$k + 8 = 0$
$k = -8$
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