How do you determine the remaining zeros for $h(x)=3{{x}^{4}}+5{{x}^{3}}+25{{x}^{2}}+45x-18$ if 3i is a zero?

Hint: In this question, we have to find the remaining zeroes of a biquadratic equation. Thus, we will use the complex number property, the long division method, and splitting the middle term method. First, we will apply the complex number property, which states that if iota is one of the roots of a polynomial, then its conjugate is also the root of the same polynomial. Thus, we will apply the same property in 3i root and thus, get another root of the equation. Then, we will apply the algebraic identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ in both the factors of the polynomial, to get a quadratic polynomial. After that, we will divide the given polynomial by the quadratic polynomial. In the last, we will use splitting the middle term method in the quotient, to get the remaining zeros of the biquadratic polynomial.

Complete step-by-step solution:
According to the question, we have to find the zeros of a biquadratic polynomial.
Thus, we will use the complex number property, the long division method, and splitting the middle term method to get the solution.
The biquadratic polynomial given to us is $h(x)=3{{x}^{4}}+5{{x}^{3}}+25{{x}^{2}}+45x-18$ ------ (1)
Now, it is given to us that 3i is one of the roots of equation (1). Thus, we will apply the complex number property that states if a + bi is a root of the polynomial, where a and b are the real numbers, then its complex conjugate a − bi is also a root of the same polynomial. Therefore, the another root of equation (1) is -3i, therefore the factors are
$\Rightarrow \left( x-3i \right)\left( x+3i \right)$
Now, we will apply the algebraic identity $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ in the above expression, we get
$\Rightarrow {{\left( x \right)}^{2}}-{{\left( 3i \right)}^{2}}$
On further solving the above expression, we get
$\Rightarrow {{x}^{2}}+9$ ------- (2)
Thus, we see the above expression is a quadratic polynomial. Therefore, we will divide the equation (1) by the quadratic equation (2) using the long-division method, we get
\[\begin{align}
  & \text{ 3}{{\text{x}}^{2}}+5x-2 \\
 & {{x}^{2}}+9\left| \!{\overline {\,
 3{{x}^{4}}+5{{x}^{3}}+25{{x}^{2}}+45x-18 \,}} \right. \\
 & \text{ }\underline{{}_{-}\text{3}{{\text{x}}^{4}}+0{{x}^{3}}+{}_{-}27{{x}^{2}}} \\
 & \text{ 5}{{\text{x}}^{3}}-2{{x}^{2}}+45x-18 \\
 & \text{ }\underline{{}_{-}\text{5}{{\text{x}}^{3}}+{}_{-}0{{x}^{2}}+{}_{-}45x}\text{ } \\
 & \text{ -2}{{\text{x}}^{2}}\text{ }-18 \\
 & \text{ }\underline{\text{ -}{}_{+}\text{2}{{\text{x}}^{2}}-{}_{+}18} \\
 & \text{ }\underline{\text{ 0 }} \\
 & \text{ } \\
\end{align}\]
Thus, we get the remainder equals to 0 and a quotient equal to $3{{x}^{2}}+5x-2$ .
Now, we will use the splitting the middle term method in the quotient, which is
$3{{x}^{2}}+5x-2$
So, we will split the middle term as the sum of 6x and -x, because $a.c=6.\left( -1 \right)=-6$ and $a+c=6-1=5$
Therefore, we get
$\Rightarrow 3{{x}^{2}}+6x-x-2$
Now, we take 3x common from the first two terms and take common -1 from the last two terms, we get
$\Rightarrow 3x(x+2)-1(x+2)$
Now, take (x+2) common from the above equation, we get
$\Rightarrow (x+2)(3x-1)$
So, we will solve either
$\Rightarrow (x+2)=0$ ------ (3) and
$\Rightarrow (3x-1)=0$ -------- (4)
Now, we will first solve equation (3), by subtracting 2 on both sides of the equation, we get
$\Rightarrow x+2-2=0-2$
As we know, the same terms with opposite signs cancel out each other, thus we get
 $\Rightarrow x=-2$
Now, we will first solve equation (4), by adding 1 on both sides of the equation, we get
$\Rightarrow 3x-1+1=0+1$
As we know, the same terms with opposite signs cancel out each other, thus we get
 $\Rightarrow 3x=1$
So, we will divide 3 on both sides in the above equation, we get
$\Rightarrow \dfrac{3}{3}x=\dfrac{1}{3}$
On further simplification, we get
$\Rightarrow x=\dfrac{1}{3}$
Therefore, for the equation $h(x)=3{{x}^{4}}+5{{x}^{3}}+25{{x}^{2}}+45x-18$ , the value of zeroes, that is value of x is equal to $-2,\dfrac{1}{3},3i,-3i$.

Note: While solving this problem, do mention all the formula and the property you are using to avoid confusion and mathematical error. To solve the quotient after the long division method, you can also use the cross multiplication method to get the solution.