\( x^{3}\)+\(6x^{2} \)+\( 11x\)+\( 6\) is divisibke by
Correct Answer: Description for Correct answer:
From option (a)
x + 1 = 0 = > x = -1
For x = -1, \( \Large x^{3} + 6x^{2} + 11x + 6 \)
\( \Large = (-1)^{3} + 6(-1)^{2} + 11(-1) + 6 \)
= -1 + 6 - 11 + 6 = 0
Hence \( \Large x^{3} + 6x^{2} + 11x + 6 \) is divisible by x + 1.
From option (b)
x + 2 = 0 => x = -2
For x = -2
\( \Large = x^{3} + 6x^{2} + 11x + 6 \)
\( \Large = (-2)^{3} + 6(-2)^{2} + 11(-2) + 6 \)
= -8 + 24 - 22 + 6
= 30 - 30 = 0
Hence \( \Large x^{3} + 6x^{2} + 11x + 6 \) is also divisible by x + 2.
From option c
x + 3 = 0 = > x = -3
For x = -3,
\( \Large x^{3} + 6x^{2} + 11x +b \)
\( \Large = (-3)^{3} + 6(-3)^{2} + 11(-3) + 6 \)
= -27 + 54 - 33 + 6
= 60 - 60 = 0
Hence \( \Large x^{3} + 6x^{2} + 11x + 6 \) is also divisible by x + 3 = 0
Hence \( \Large x^{3} + 6x^{2} + 11x + 6 \) is also divisible by
(x + 1), (x + 2) and (x + 3)
option (d) is correct.
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