Question

If k is a constant, what is the value of k such that the polynomial 2ky^-kx-8 is divisible by (x-2)

Answer 1

Answer:we have that

k² x³-6kx+9

we know that

The Remainder Theorem says: If the polynomial f(x) is divisible by a bynomial (x-a) then the number "a" is a root of the polynomial f(x)

so

x=1  is a root of the polynomial

therefore

for x=1

k² x³-6kx+9=0--------> k² (1)³-6k(1)+9=0-----> k²-6k+9=0

using a graph tool------> to resolve the second order equation

the solution is

k=3

k=3

k²-6k+9=0---------> (k-3)²=0

the answer is

k=3

Answer 2

Answer:

Ok so if I did this right the answer will be  2ky^kx-8 / x   -2