Chapter 4.3, Problem 32AYU

To determine

To find: The Complex zeros of the polynomial function $f\left(x\right)$ whose coefficients are real numbers and to write in factored form.

Answer to Problem 32AYU

$\text{Factored form of }f\left(x\right)=x^4-1=\left(x+1\right)\left(x-1\right)\left[x-i\right]\left[x+i\right]$.

Explanation of Solution

Given:

$f\left(x\right)=x^4-1$

Degree of the polynomial function $f\left(x\right)=x^4-1=4$.

Therefore $f\left(x\right)$ have 4 zeros.

From Descartes’ rule of signs,

$f\left(x\right)=x^4-1$

$+$ to $-$

Therefore $f\left(x\right)=x^4-1$ have 1 positive real zero.

$f\left(-x\right)=x^4-1$

$+$ to $-$

Therefore $f\left(x\right)=x^4-1$ have one negative real zero.

$f\left(x\right)=x^4-1$ in factored form $=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)$.

$\text{The complex zero }=x^2= - 1$.

$x= \pm i$

$\text{The complex zeros }=x=\pm i$.

Factored form of $f\left(x\right)=x^4-1$.

$=\left(x+1\right)\left(x-1\right)\left[x-i\right]\left[x-\left(-i\right)\right]$

$=\left(x+1\right)\left(x-1\right)\left[x-i\right]\left[x+i\right]$