How to recognize the natural k order of multiplicity?the root x=1 is multiple in `6x^7-7x^6+1=0` how many times is multiple?
You need to find the order of multiplicity of the given root, hence, you need to remember that the root of order `k` of multiplicity is not only the root of polyinomial but it is also the root of all derivatives till `(k-1)th`derivative .
Hence, you need to find the first derivative such that:
`f'(x) = (6x^7-7x^6+1)' = 42x^6 - 42x^5`
Chacking if `x = 1` is the root to `f'(x)` yields:
`f'(1) = 42(1^6 - 1^5) = 0 => x = 1 ` is the root of `f'(x)`
You need to find the second derivative such that:
`f''(x) = 252x^5 - 210x^4 `
You need to substitute 1 for x in equation of second derivative such that:
`f''(1) = 252 - 210 != 0`
Hence, evaluating the order of multiplicity k of the given root `x = 1` , yields that `k = 2` .