$If x^{3} + 6 x^{2} + 4 x + k is divisible by (x + 2), then the value of k is:$
-8

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Solution

The correct option is A -8
$Given: Polynomial, p (x) = x^{3} + 6 x^{2} + 4 x + k (x + 2) is a factor of p (x) .$

$By factor theorem, if (x - a) is a factor of p (x), then p (a) = 0.$

$Using factor theorem, p (- 2) = 0 {(- 2)}^{3} + 6 \times {(- 2)}^{2} + 4 (- 2) + k = 0 - 8 + 24 - 8 + k = 0 k = - 8$

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If x^{3} + 6 x^{2} + 4 x + k is divisible by (x + 2), then the value of k is:

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(i) $2 x^{3} - 3 x^{2} - 17 x + 30$

(ii) $x^{3} - 6 x^{2} + 11 x - 6$

(iii) $x^{3} + x^{2} - 4 x - 4$

(iv) $3 x^{3} - x^{2} - 3 x + 1$

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If x3+6x2+4x+k is divisible by (x+2),then the value of k is:-8

The correct option is A -8Given: Polynomial, p(x)=x3+6x2+4x+k(x+2) is a factor of p(x). By factor theorem, if (x−a)is a factor of p(x),then p(a)=0. Using factor theorem,p(−2)=0(−2)3+6×(−2)2+4(−2)+k=0−8+24−8+k=0k=−8

$If x^{3} + 6 x^{2} + 4 x + k is divisible by (x + 2), then the value of k is:$
-8