Question

How to find all the zeros of ${x^4} - 4{x^3} + 14{x^2} - 4x + 13$ with the zero $2 - 3i$?

Answer 1

In this question, we are given with an equation ${x^4} - 4{x^3} + 14{x^2} - 4x + 13$ which has a zero $2 - 3i$ and we need to find all the other zeros. We know that complex zeros always occur as conjugates. If a zero $c$ is given, to find the rest of zeros, we will divide the given polynomial by the polynomial $(x - c)$ . In this way we will find the other zeroes.

Formula used: If a zero $c$ is given, to find the rest of zeros, we will divide the given polynomial by $(x - c)$

Complete step by step solution:
In this question, we have to find all the zeros of ${x^4} - 4{x^3} + 14{x^2} - 4x + 13$ with the zero $2 - 3i$.
It is given that the function ${x^4} - 4{x^3} + 14{x^2} - 4x + 13$ has a zero of $2 - 3i$, and all the coefficients of the given polynomial equation ${x^4} - 4{x^3} + 14{x^2} - 4x + 13$ are real.
We know that, when a polynomial has complex zero, then its conjugate is also a zero of the same polynomial.
Thus $2 - 3i$ is zero of the given polynomial then its conjugate $2 + 3i$ is also zero in the given polynomial equation.
Therefore $2 - 3i$ and $2 + 3i$are zeros of the given polynomial equation.
Since $2 - 3i$ and $2 + 3i$are roots of the given polynomial equation,
We can use it to find the other two zeroes.
If a zero $c$ is given, to find the rest of zeros, we will divide the given polynomial by $(x - c)$
Now, we know that two of the polynomial's factors are $(x - (2 - 3i))\;$ and $(x - (2 + 3i)).$
Multiplying it, we get
$\Rightarrow (x - (2 - 3i))\;(x - (2 + 3i))$
On rewriting we get,
$\Rightarrow (x - 2 + 3i)(x - 2 - 3i)\;$
On splitting we get,
$\Rightarrow ((x - 2) + 3i)((x - 2) - 3i)$
On taking the term and we get,
$\Rightarrow {(x - 2)^2} - {(3i)^2}$
On taking the square and we get,
$\Rightarrow {x^2} - 4x + 4 + 9$
Let us add the term and we get,
$\Rightarrow {x^2} - 4x + 13$
The remaining factors of the polynomial can be found by dividing the given polynomial equation by this resultant equation.
$\Rightarrow \dfrac{{{x^4} - 4{x^3} + 14{x^2} - 4x + 13}}{{{x^2} - 4x + 13}}$
Splitting $14{x^2}$ into$13{x^2} + 1{x^2}$ , we get
$\Rightarrow \dfrac{{{x^4} - 4{x^3} + 13{x^2} + 1{x^2} - 4x + 13}}{{{x^2} - 4x + 13}}$
Taking out the commons,
$\Rightarrow \dfrac{{{x^2}\left( {{x^2} - 4x + 13} \right) + 1\left( {{x^2} - 4x + 13} \right)}}{{{x^2} - 4x + 13}}$
Again, on taking common term and we get,
$\Rightarrow \dfrac{{\left( {{x^2} - 4x + 13} \right)\left( {{x^2} + 1} \right)}}{{{x^2} - 4x + 13}}$
Cancel the term and we get
$\Rightarrow {x^2} + 1$
Therefore $\dfrac{{{x^4} - 4{x^3} + 14{x^2} - 4x + 13}}{{{x^2} - 4x + 13}} = {x^2} + 1$
Thus the remaining zeros can be solved by as follows,
${x^2} + 1 = 0$
${x^2} = - 1$
$x = \sqrt { - 1}$
$x = \pm i$
The function has four imaginary zeros and never crosses the x-axis.

Therefore the zeroes are: $- {\text{ }}i{\text{ }},{\text{ }}i{\text{ }},{\text{ }}2{\text{ }} + {\text{ }}3{\text{ }}i,{\text{ }}2{\text{ }} - {\text{ }}3{\text{ }}i.$

Note: students may get confused or may go wrong in understanding what conjugate or conjugation means.
In Algebra, the conjugation is a method where you should change the signs (+ to −, or − to +) between two terms.
For example: $3x{\text{ }} + {\text{ }}1\; \to 3x{\text{ }} - {\text{ }}1$ ,where the sign in the middle of the given two terms is changed.
Another example is $2z{\text{ }} - {\text{ }}7{\text{ }} \to {\text{ }}2z{\text{ }} + {\text{ }}7$ and $5i - 1 \to 5i + 1$