Chapter 4, Problem 78RE

To determine

To find: the remaining zeros off and polynomial function

Answer to Problem 78RE

  $1-i$ must also be the zeros of $f(x)$ .

The polynomial $f(x)=x^4-5x^3+10x^2-10x+4$

Explanation of Solution

Given:

Degree $4;$ zeros: $1,2,1+i$ .

Calculation:

The conjugate pair theorem states that if $a+bi$ is a complex zero of a polynomial with real coefficients, then its complex conjugate $a-bi$ is also a zero of the polynomial.

Since the polynomial has a degree 4, the polynomial will have four zeros.

Thus, $1-i$ must also be the zeros of $f(x)$ .

Find the factors of $f(x)$ -

By the factor theorem, if $f(c)=0,$ then $x-c$ will be a factor of $f(x)$ .

Thus, $f(x)=a(x-1)(x-2)[x-(1+n)][x-(1-n)]$ . where $a$ is any real number.

Remove the parentheses using the distributive property and simplify.

  $\begin{array}{l}f(x)=a(x-1)(x-2)[x-(1+i)][x-(1-i)]\\ =a\left(x^2-2x-x+2\right)\left[x^2-x(1-i)-x(1+i)+(1+i)(1-i)\right]\\ =a\left(x^2-3x+2\right)\left[x^2-x+xi-x-xi+1-i^2\right]\\ =a\left(x^2-3x+2\right)\left[x^2-2x+1+1\right]\\ =a\left(x^2-3x+2\right)\left[x^2-2x+2\right]\\ =a\left(x^4-2x^3+2x^2-3x^3+6x^2-6x+2x^2-4x+4\right)\\ =a\left(x^4-5x^3+10x^2-10x+4\right)\end{array}$

Let $a=1.$

Thus, the polynomial $f(x)=x^4-5x^3+10x^2-10x+4$

Conclusion:

Thus, $1-i$ must also be the zeros of $f(x)$ .

Thus, the polynomial $f(x)=x^4-5x^3+10x^2-10x+4$