since complex roots come in pairs, the factors are
(x-5)(x-i)(x+i)(x-(-6+i))(x-(-6-i))
(x-5)(x-i)(x+i)((x-6)-i)((x-6)+i)
(x-5)(x^2+1)(x^2+12x+37)
form a polynomial f (x) with real coefficients having the given degree and zeros.
degree 5; zeros 5; -i; -6+-i
f(x)= a(?)
Please show step by step work.
1 answer