Question

if x cube add 8x square add kx add 198 is compared divisible by x square add 6x add 9 , then find the value of k?​

Answer 1

Step-by-step explanation:

Given that

${x}^3 + 8 {x}^{2} + kx + 198$

is completely divisible by

${x}^{2} + 6x + 9$

We can write

${x}^{2} + 6x + 9 = {(x + 3)}^{2}$

So (x+3) is a factor of the polynomial

${x}^{3} + 8 {x}^{2} + kx + 198$

Hence when we substitute x= -3 in this polynomial we will get zero.

${( - 3)}^{3} + 8 {( - 3)}^{2} +k( - 3) + 198 = 0$

$= > - 27 + 72 - 3k + 198 = 0 = >$

243+3k=0 =>k =(-243)/3= -81. Hope it helps you.