Question

If x^3+8x^2+kx+18 is completely divisible by x^2+6x+9,then find the value of k

Answer 1

x^3 + 8x^2 + kx + 18 is completely divisible by x^2 + 6x + 9
first of all we should break x^2 + 6x + 9
e.g., x^2 + 2.3x + 3^2 = (x + 3)^2 
hence, x = -3 is the zero 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 

hence, k = 21

Answer 2

Given ⇒

 f(x) = x³ + 8x² + kx = 18

Divisor of the Polynomial = x² + 6x + 9
∴ x² + 6x + 9 = 0
Splitting the middle term,
x² + 3x + 3x + 9 = 0
x(x + 3) + 3(x + 3) = 0
(x + 3)(x + 3) = 0
(x + 3)² = 0
∴ x + 3 = 0
∴ x = -3

Now, (x + 3) can divide the polynomial f(x) completely.
∴ Remainder f(x) = 0
∴ 0 = (-3)³ + 8(-3)² + k(-3) + 18
∴ 0 = -27 + 8(9) -3k + 18
∴ 72 - 27 - 3k + 18 = 0
⇒ 3k  = 45 + 18
⇒ k = 63/3
∴ k = 21

Hence, the value of the k is 21.


Hope it helps.