math
posted by april .
use the given zero to find the remaining zeros of the fuction...f(x)=x to the 5power6x4power+7x3power10x2power 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 1i
so (x (1+i)) and (1  (1i)) are factors
expanded gives us:
(x1i)(x1+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 + 7x10)
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 , 1i