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
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