Chapter 4.3, Problem 31AYU

In Problems 31-40, find the complex zeros of each polynomial function. Write $f$ in factored form.

$f\left(x\right)=x^3-1$

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 31AYU

$\text{Factored form of }f\left(x\right)=x^3-1=\left(x-1\right)\left[x-\left(\frac{-1+i\sqrt{3}}{2}\right)\right]\left[x-\left(\frac{-1-i\sqrt{3}}{2}\right)\right]$

Explanation of Solution

Given:

$f\left(x\right)=x^3-1$

Degree of the polynomial function $f\left(x\right)=x^3-1=3$.

Therefore $f\left(x\right)$ have 3 zeros.

From Descartes’ rule of signs,

$f\left(x\right)=x^3-1$

$+$ to $-$

Therefore $f\left(x\right)=x^3-1$ have 1 positive real zero.

$f\left(x\right)=-x^3-1$

Therefore $f\left(x\right)=x^3-1$ have no negative real zeros.

By Synthetic division,

The coefficients of $f\left(x\right)\text{ = 1, 0, 0, }-1$.

The positive real zero $\text{= 1}$.

The depressed equation $=x^2+x+1$.

The quadratic equation $=a{x}^2+bx+c$.

$a\text{ = 1}$, $b\text{ = 1}$, $c\text{ = 1}$.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-1\pm \sqrt{1^2-\mathrm{4.1.1}}}{2.1}$

$x=\frac{-1\pm \sqrt{-3}}{2}=\frac{-1\pm i\sqrt{3}}{2}$

The complex zeros $=\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}$.

Factored form of $f\left(x\right)=x^3-1$

$=\left(x-1\right)\left[x-\left(\frac{-1+i\sqrt{3}}{2}\right)\right]\left[x-\left(\frac{-1-i\sqrt{3}}{2}\right)\right]$