Chapter 4, Problem 77RE

To determine

To calculate: Remaining zeros of $f$ .

Answer to Problem 77RE

Thus answer is $f\left(x\right)=x^4-2x^3+3x^2-2x+2$ .

Explanation of Solution

Given information:

Degree: $4$

Zeros: $i,1+i$

Calculation:

If the given zeros of a polynomial function has non real complex numbers then the other root is the conjugate the complex number.

So if the zeros are $i$ and $1+i$ , then the other zeros are $-i$ and $1-i$ .

To determine the factors of the polynomial set the variable of the polynomial equal to zeros of $f\left(x\right)$ .

For determining the factors of the polynomial set the variable of the polynomial equal to the zeros of $f\left(x\right)$ .

  $\begin{array}{l}x=i\\ x=-i\\ x=1+i\\ x=1-i\end{array}$

Perform opposite operation to make the right side zero.

  $\begin{array}{l}x-i=0\\ x+i=0\\ x-1-i=0\\ x-1+i=0\end{array}$

So the factors of $f\left(x\right)$ are $x-1,x+i,x-1-i$ and $x-1+i$ .

Then multiply the factors,

  $\begin{array}{l}\left(x-i\right)\left(x+i\right)\left(x-1-i\right)\left(x-1+i\right)\\ =\left(x^2+ix-ix-i^2\right)\left(x^2-x+ix-x+1-i-ix+i-i^2\right)\\ =\left(x^2-i^2\right)\left(x^2-2x+1-i^2\right)\end{array}$

Note that $i^2-1$

  $\begin{array}{l}=\left(x^2+1\right)\left(x^2-2x+1-\left(-1\right)\right)\\ =\left(x^2+1\right)\left(x^2-2x+2\right)\\ =x^4-2x^3+2x^2+x^2-2x+2\\ =x^4-2x^3+3x^3-2x+2\\ =x^4-2x^3+3x^2-2x+2\end{array}$