Question

If the expression for the function is given by $f(x)$ $f(x)={{x}^{4}}-4{{x}^{3}}+4{{x}^{2}}+8x+44$, then find \[f\left( 3+2i \right)\].

Answer 1

Hint: Here, we are given the value of x and f(x), therefore, we can use the given value of x in the expression for f(x) to obtain the required answer.

Complete step-by-step answer:
It is given that $f(x)={{x}^{4}}-4{{x}^{3}}+4{{x}^{2}}+8x+44$. Also, the value of x for which f(x) is to be evaluated is also given. We note that $i=\sqrt{-1}$. Therefore,
${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$, ${{i}^{3}}={{\left( \sqrt{-1} \right)}^{3}}=-\sqrt{-1}=-i$ and ${{i}^{4}}={{\left( \sqrt{-1} \right)}^{4}}=1............................(1.1)$
We also know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}........(1.2)$. Therefore, we can obtain the powers of the given value of x as
$\begin{align}
  & x=3+2i \\
 & \Rightarrow {{x}^{2}}={{\left( 3+2i \right)}^{2}}={{3}^{2}}+2\times 3\times 2i+{{2}^{2}}\times {{i}^{2}} \\
 & =9+12i-4\text{ }(\text{as }{{i}^{2}}\text{=-1 form equation 1}\text{.1}) \\
 & =5+12i...........................(1.3) \\
\end{align}$
Also, we can obtain the cube of x by multiplying the square of x by x. Therefore,
$\begin{align}
  & {{x}^{3}}={{x}^{2}}\times x=(5+12i)(3+2i)\text{ (from 1}\text{.3)} \\
 & \text{=5}\times \text{3+5}\times 2i+12\times \text{3i+12}\times 2{{i}^{2}} \\
 & =15+46i-24\text{ (using equation 1}\text{.1)} \\
 & \text{=}-9+46i...................(1.3) \\
\end{align}$
And the fourth power of x can be obtained by taking the square of ${{x}^{2}}$. Therefore,
$\begin{align}
  & {{x}^{4}}={{\left( {{x}^{2}} \right)}^{2}}={{\left( 5+12i \right)}^{2}}={{5}^{2}}+2\times 5\times 12i+{{12}^{2}}{{i}^{2}} \\
 & =25+120i-144\text{ (as }{{i}^{2}}\text{=}-1\text{ from 1}\text{.1)} \\
 & \text{=}-119+120i....................(1.4) \\
\end{align}$
Therefore, from the equations (1.2), (1.3) and (1.4) and the given value of x, we can evaluate f(x) at x=3+2i as
$\begin{align}
  & f(3+2i)={{(3+2i)}^{4}}-4{{(3+2i)}^{3}}+4{{(3+2i)}^{2}}+8(3+2i)+44 \\
 & =\left( -119+120i \right)-4(-9+46i)+4(5+12i)+8(3+2i)+44 \\
 & =\left( -119+4\times 9+4\times 5+8\times 3+44 \right)+\left( 120-4\times 46+4\times 12+8\times 2 \right)i \\
 & =5+0i=5 \\
\end{align}$

Therefore, we obtain the answer to the given question to be 5.

Note: In the solution, we directly evaluated the powers of x and put it in the expression for f(x). However, we could also have found the quadratic equation which evaluates to zero from the given value of x, for example in this case, x=3+2i. Therefore, we have
$\begin{align}
  & x-3=2i\Rightarrow {{\left( x-3 \right)}^{2}}={{2}^{2}}{{i}^{2}}=-4 \\
 & \Rightarrow {{x}^{2}}-6x+9=-4\Rightarrow {{x}^{2}}-6x+13=0 \\
\end{align}$
Therefore, we can try to divide the polynomial f(x) by the polynomial ${{x}^{2}}-6x+13$. Then the term which is a factor of ${{x}^{2}}-6x+13$ will evaluate to zero when the value of x is put and the value of the remainder evaluated at x will give us the required answer to the question.