Question

Find the value of ${{x}^{4}}-{{x}^{3}}+{{x}^{2}}+3x-5$ if $x=2+3i$.

Answer 1

Hint: Solve each term separately then substitute everything back into the equation. By using the distributive law a(b + c) = a.b + a.c

Complete step-by-step solution -
Given expression in the question:
${{x}^{4}}-{{x}^{3}}+{{x}^{2}}+3x-5$
Given value of $x$ in the question is: $x=2+3i$
By substituting the value of $x$ into expression we get:
${{x}^{4}}-{{x}^{3}}+{{x}^{2}}+3x-5={{\left( 2+3i \right)}^{4}}-{{\left( 2+3i \right)}^{3}}+{{\left( 2+3i \right)}^{2}}+3\left( 2+3i \right)-5$
Now we will separate the equation into 4 parts.
The 4 parts be named as A, B, C, D
Let
$\begin{align}
  & A=3\left( 2+3i \right)-5 \\
 & B={{\left( 2+3i \right)}^{2}} \\
 & C={{\left( 2+3i \right)}^{3}} \\
 & D={{\left( 2+3i \right)}^{4}} \\
\end{align}$
We should solve each term separately
By solving A: $A=3\left( 2+3i \right)-5$
By using distributive law:$ a(b + c) = a.b + a.c$
A = $3.2 +3.3i – 5$
By simplifying we get:
A = $1 + 9i $
By solving B: $B={{\left( 2+3i \right)}^{2}}$
We can write B as: $B=\left( 2+3i \right)\left( 2+3i \right)$
By applying distributive law: a(b + c) = a.b + a.c
By using distributive law twice, we get:
B = $2.2 + 2.3i + 3i.2 + 3i.3i$
By simplifying we get:
B = $-5 + 12i$
By solving C: $C={{\left( 2+3i \right)}^{3}}$
We can write C as: $C=\left( 2+3i \right)\left( 2+3i \right)\left( 2+3i \right)$
 By applying distributive law: a(b + c) = a.b + a.c
By using distributive law thrice, we get:
C =$ 2.2.2 + 3.2.3i.2 + 3.2.3i.3i + 3i.3i.3i$
By simplifying we get:
C = $-46 + 9i$
By solving D: $D={{\left( 2+3i \right)}^{4}}$
 We can write D as: $D=\left( 2+3i \right)\left( 2+3i \right)\left( 2+3i \right)\left( 2+3i \right)$
By applying distributive law: a(b + c) = a.b + a.c
By applying distributive law four times, we get:
D = $2.2.2.2 + 3i.3i.3i.3i + 4.4.3i.3i + …….$
D = $-119 – 120i$
By substituting A, B, C, D back into equation, we get:
$\begin{align}
  & {{x}^{4}}-{{x}^{3}}+{{x}^{2}}+3x-5=\left( -119-120i \right)+46-9i+\left( -5+12i \right)+1+9i \\
 & =-119-120i+46-9i-5+12i+1+9i
\end{align}$
By grouping real and imaginary terms separately, we get:
${{x}^{4}}-{{x}^{3}}+{{x}^{2}}+3x-5=-119+46-5+1-120i-9i+12i+9i$
By simplifying, we get:
${{x}^{4}}-{{x}^{3}}+{{x}^{2}}+3x-5=-119+46-5+1-108i$
By simplifying real part, we get:
${{x}^{4}}-{{x}^{3}}+{{x}^{2}}+3x-5=-77-108i$
Therefore, the value of the given expression at $x=2+3i$ is $-77-108i$ .

Note: Be careful while calculating 4th degree terms because you have to apply distributive law many times you may make a mistake there.
Alternative method is to use algebraic identity
${{\left( a+b \right)}^{4}}={{a}^{4}}+6{{a}^{2}}{{b}^{2}}+{{b}^{4}}+4{{a}^{3}}b+4a{{b}^{3}}$