Question

if x cube + 8 x square + kx + 18 is completely divisible by X square + 6 X + 9 find the value of k​

Answer 1

Answer:

x^3 + 8x^2 + kx + 18 is completely divisible by x^2 + 6x + 9

firstly,  break x^2 + 6x + 9

e.g., x^2 + 2.3 x + 3^2 = (x + 3)*( x + 3)

hence, x = -3 is the root of x^3 + 8x^2 + kx + 18

now, put x = -3 in  x^3 + 8x^2 + kx + 18 

(-3)^3 + 8(-3)^2 + k(-3) + 18 = 0

⇒ -27 + 8 × 9 -3k + 18 =0

⇒ -27 + 72 - 3k + 18 = 0

⇒ 63 - 3k = 0

⇒  k = 63/3 = 21 

Answer 2

Answer:

K=63/3 =21...........