Question

If x^3+6x^2+4x+k is exactly divisible by x+2, then k =
(a) - 6
(b) - 7
(c) - 8
(d) -10​

Answer 1

Answer:

C

Step-by-step explanation:

Correct option is

C

-8

Divide x

3

+6x

2

+4x+k by x+2

x+2

)x

3

+6x

2

+4x+k(

x

2

+4x−6

−(x

3

+2x

2

)

4x

2

+4x+k

−(4x

2

+8x)

−4x+k

−(−4x−8)

k+8

As, x+2 is a factor of the given polynomial, so remainder must be 0.

So, k+8=0 or k=−8

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Hence, C is correct.

Answer 2

$\large \mathfrak \green {➯ \: Question \: ࿐}$

If $\sf x^3+6x^2+4x+k$ is divisible exactly by $\sf x+2$, then k is equal to :-

(A) -6

(B) -7

(C) -8

(D) -10

$\large \mathfrak \green {➯ \: Solution \: ࿐ }$

It is given that $\sf x+2$ divides $\sf x^3+6x^2+4x+k$. So :-

$\pink {:\implies} \: \: \sf x+2=0$

$\pink {:\implies} \: \: \sf x= - 2$

Now, substituting $\sf x=2$ in the equation and equating it to 0 :-

$\pink {:\implies} \: \: \sf (-2)^3+6(-2)^2 + 4( - 2) + k = 0$

$\pink {:\implies}\:\:\:\sf -8+6(4) - 8 + k = 0$

$\pink {:\implies}\:\:\:\sf -8+24 - 8 + k = 0$

$\pink {:\implies} \: \: \: \sf -16+24+k=0$

$\pink {:\implies} \: \: \: \sf 8+k=0$

$\pink {:\implies} \: \: \: \sf k=-8$

$\pink {\therefore\sf \: k=-8(OptionC) \: when \: x+2 \: divides \: x^3+6x^2+4x+k=0.}$

$\large \mathfrak \green {➯ \: Verification \: ࿐}$

Substituting $\sf k=-8$ and $\sf x=-2$ in the equation and equating it to 0 :-

$\pink {:\implies}\:\:\sf(-2)^3+6(-2)^2+4(-2)+(-8)=0$

$\pink {:\implies} \: \: \: \sf -8 + 6(4) - 8 - 8 = 0$

$\pink {:\implies} \: \: \: \sf -24+24=0$

$\pink {:\implies} \: \: \: \sf 0=0$

$\pink {\therefore \sf \:LHS=RHS \: }$

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