Question

if x square 3 + 6x square 2 + kx +6 is exactly divisible by (x+2) then k= ?​

Answer 1

Answer:

 If $x^3 + 6x^2 + kx + 6$ is exactly divisible by (x+2)  then k is 11.

Step-by-step explanation:

If $x^3 + 6x^2 + kx + 6$ is exactly divisible by (x+2)

then put x = -2 in $x^3 + 6x^2 + kx + 6$

$x^3 + 6x^2 + kx + 6 = 0\\(-2)^3 + 6(-2)^2 + k(-2) + 6 =0\\-8 + 6(4) - 2k + 6 =0\\-8 + 24 -2k+ 6 =0\\22 - 2k = 0\\2k = 22\\k = 22/2\\ k = 11$

So,  If $x^3 + 6x^2 + kx + 6$ is exactly divisible by (x+2)  then k is 11.