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in order to have real coefficients, all complex roots must appear as conjugates
do you know what the conjugate of 2+6i is?
http://www.mathsisfun.com/numbers/complex-numbers.html a+bi and a-bi are complex conjugates
yep so can you write the polynomial in factored form?
7, -11, and 2 + 6i
no... in factored form it would be something like this... f(x) = (x-c)(x-d)(x-(a+bi))(x-(a-bi))
f(x)=(x+7)(x -11) (2 + 6i)
not quite... the complex factor would be x-(2+6i) and you would need its conjugate as a factor also, if your polynomial is to have real coefficients
i don't really understand
if 4 is a root, it implies that f(4) = 0 and that the polynomial must have a factor of x-4, correct?
so if 2+6i is a root, then x-(2+6i) is a factor. but if your polynomial is to have real coefficients then all complex roots must occur as complex conjugate pairs
so x-(2-6i) also must be a factor
once you have the factored form, just multiply out to get into standard form