Complex roots of polynomial equations with real coefficients Consider the polynomial $x^5 +ax^4 +bx^3 +cx^2 +dx+4$ where $a, b, c, d$ are real
numbers. If $(1 + 2i)$ and $(3 - 2i)$ are two roots of this polynomial then what is the value of $a$ ?
 A: As the coefficients of the different powers of $x$ are real, all the complex roots must occur with conjugate pair.
So, the other two roots are $1-2i,3+2i$ 
So if $p$ is the fifth root,
using Vieta's Formulas $(1-2i)(3+2i)(1+2i)(3-2i)p=(-1)^5\frac41$
and $(1-2i)+(3+2i)+(1+2i)+(3-2i)+p=-\frac a1$
A: Adding to lab bhattacharjee's answer, Vieta's Formulas basically tell you that the negative fraction of the last term (the constant) divided by the coefficient of the first term is equal to the product of the roots. Letting r be the 5th root of the polynomial (since we know 4), $$-\frac{4}{1} = (1−2i)(1+2i)(3+2i)(3-2i)r$$  We get $1+2i$ and $3-2i$ as two other roots because they're conjugates of the roots you gave us. Roots as well as their conjugates are roots of a  polynomial. Vietas formulas also states that the negtive fraction of the coefficient of the term immediately after the first term divided by the coefficient of the first term is equal to the sum of the roots. So $$-\frac{a}{1} = (1−2i)+(1+2i)+(3+2i)+(3-2i)+r$$ Using these two equations, you can solve for $a$ and $r$.
