use the given zero to find the remaining zeros of the fuction...f(x)=x to the 5power-6x4power+7x3power-10x2power zero:1+i
type it this way:
f(x) = x^5 - 6x^4 + 7x^3 - 10x^2
since complex number zeros always come in conjugate pairs, there is another zero of 1-i
so (x -(1+i)) and (1 - (1-i)) are factors
expanded gives us:
(x-1-i)(x-1+i)
= x^2 - x + ix - x + 1 - i - ix + i - i^2 , but i^2 = -1
= x^2 - 2x + 2
so we have
f(x) = x^2(x^3 - 6x^2 + 7x-10)
but we know that (x^2 - 2x + 2) must be a factor of
(x^3 - 6x^2 + 7x - 10)
that is:
( x^2 - 2x + 2)(......) = x^3 - 6x^2 + 7x - 10
that would leave a binomial factor of (x - 5)
so the zeros are
x = 0 , 5 , 1+i , 1-i