.If (x-2)is a factor of     f(x)=x³+x³-16x+k, then f(x)is     divisible by*x+1x-1x-5x+5

The question seems to be a bit unclear, but I'll try to interpret it as best as I can.

If (x-2) is a factor of f(x) = 2x³ - 16x + k, then by the Factor Theorem, f(2) = 0.

Let's substitute x = 2 into the equation:

f(2) = 2*(2)³ - 16*(2) + k = 0 16*2 - 32 + k = 0 32 - 32 + k = 0 k = 0

So, the polynomial becomes f(x) = 2x³ - 16x.

Now, to check if f(x) is divisible by x+1, x-1, x-5, or x+5, we substitute these values into the equation and see if the result is zero.

For x+1: f(-1) = 2*(-1)³ - 16*(-1) = -2 + 16 = 14 ≠ 0 For x-1: f(1) = 2*(1)³ - 16*(1) = 2 - 16 = -14 ≠ 0 For x-5: f(5) = 2*(5)³ - 16*(5) = 250 - 80 = 170 ≠ 0 For x+5: f(-5) = 2*(-5)³ - 16*(-5) = -250 + 80 = -170 ≠ 0

Therefore, f(x) is not divisible by x+1, x-1, x-5, or x+5.