Real or Complex ???

my solution is more logical than algebric $P(i)=0\longrightarrow ai+b-ci=5$ note that there is only 1 real part, and the rest are complex, but the sum is real, so the sum of the complex must be zero, for which, $\quad a=c$ $ai-ci+b=5\longrightarrow b=5$ $P(2)=0\longrightarrow 35+8a+4b+2c=0$ since a=c and b=5 $35+10a+20=0\longrightarrow a= \boxed{-5.5}$

It is given that $P(z) = 2z^4+az^3+bz^2+cz+3$ and that $P(2) = 0$ and $P(i) = 0$

Therefore,

$\begin{cases} P(2) = 32+8a+4b+2c+3 = 0 & ...(1) \\ P(i) = 2-ai-b+ci+3 = 0 & ...(2) \end {cases}$

Equating the real and imaginary parts of Eqn 2, we have:

$\begin{cases} \Re : 2-b+3 = 0 \quad \Rightarrow b = 5 \\ \Im : -a+c = 0 \quad \Rightarrow a = c \end {cases}$

Substituting the values in Eqn 1, we get:

$P(2) = 32+8a+4b+2c+3 = 32 +8a + 20 +2a +3 = 0$

$\Rightarrow 10a = -55\quad \Rightarrow a = \boxed{-5.5}$

It is simple to extrapolate the two other roots from the given information. First of all, using the complex conjugate root theorem, we can easily claim that $-i$ is also a root of this polynomial.

Let $k$ be the remaining unknown root.

From Vieta's Formulas,

$2\cdot i \cdot -i \cdot k = \dfrac{3}{2} \ \ \Rightarrow \ \ k = \dfrac{3}{4}$

We now know all the roots: $2, \ i, \ -i, \ \dfrac{3}{4}$ .

Again, from Vieta's Formulas, this time the sum of roots,

$-\dfrac{a}{2} = 2 + \dfrac{3}{4} + i - i \ \ \Rightarrow \ \ \boxed{a = -\dfrac{11}{2}}$

If $a,b,c$ are real number then other factor for the polynomial is conjugate binominal of $(x-i)$ that is $x+i$ .

So

$P(Z)=(NZ+M)(Z-2)(Z-i)(Z+I)$

$P(Z)=NZ^4-(2N+M)Z^3+(2M+N)Z^2-(2N+M)Z+2M$

By the problem redaction

$NZ^4=2Z^4 \rightarrow N=2$

$2M=3 \rightarrow M=1.5$

Then $-(2N+M)=a$ $a=-(2(2)+1.5)= -5.5$

Since it is given that $a$ , $b$ and $c$ are real numbers.And one of the roots is $i$ so another root must be the complex conjugate of $i$ which is $-i$ .Let $x$ be the unknown 4th root.By Vieta's formula's: $2\times i \times -i \times x=\frac{3}{2}\\2\times1\times x=\frac{3}{2}\\2x=\frac{3}{2}\rightarrow x=\frac{3}{4}$ So our polynomial is $(Z-2)(Z-i)(Z+i)(Z-\frac{3}{4})$ .Expanding and simplifying,we get $Z^4-\frac{11}{4}Z^3+\frac{5}{2}Z^2-\frac{11}{4}Z+\frac{3}{2}=0$ .But the required polynomial has the leading coefficient as $2$ and the constant term as $3$ .So we have to multiply our polynomial by $2$ .Upon doing that,we get: $2Z^4-\frac{11}{2}Z^3+5Z^2-\frac{11}{2}Z+3\rightarrow a=-\frac{11}{2}=\boxed{-5.5}$