If f(x)=x^6–6x^4+17x^2+k, find the value of k for which x+1 is a factor of f(x). ?

f(x)=x^6–6x^4+17x^2+k

x+1 = 0

x = -1

f(x)=x^6–6x^4+17x^2+k

f(-1) = (-1)^6 – 6(-1)^4 + 17(-1)^2 + k

= 1 – 6 + 17 + k

k = -12

If f(x) = x^6 – 6x^4 + 17x^2 + k, find the value of k for which x + 1 is a factor of f(x). 

f(x) = 1 - 6 + 17 + k = 12 + k

When k has this value, another factor of f(x), of the form x + a where a is a constant

is x + 12

If (x - a) is a factor of a polynomial, then a is a 'zero' of the function.

In other words, f(a) = 0

In your case, it is asking if (x + 1) is a factor. If that is true then f(-1) = 0

Plug in x = -1 and you have:

f(-1) = (-1)^6 - 6(-1)^4 + 17(-1)^2 + k = 0

What answer do you get for k?

Figure that out and then see if you can't work out the second part as well. If not, post back with an update.

if x-a is a factor, then x=a is a root of f(x). That is, f(a) = 0. So, plug your "a" into the function and set to zero. Solve for k.

Do that, and then we can proceed .