Chapter 4.3, Problem 33AYU

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

$f\left(x\right)=x^3-8x^2+25x-26$

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 33AYU

$\text{Complex zeros }=3\pm 2i$.

$\text{Factored form of}f\left(x\right)=x^3-8x^2+25x-26=\left(x-2\right)\left[\left(x-3\right)-2i)\right]\left[\left(x-3\right)+2i)\right]$.

Explanation of Solution

Given:

$f\left(x\right)=x^3-8x^2+25x-26$

Degree of the polynomial function $f\left(x\right)=x^3-8x^2+25x-26=3$ .

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

From Descartes‘ rule of signs,

$f\left(x\right)=x^3-8x^2+25x-26$

$+$ To $-$   $-$ to $+$$+$ to $-$

Therefore $f\left(x\right)=x^3-8x^2+25x-26$ have 3 or 1 positive real zeros .

$f\left(-x\right)= -x^3-8x^2-25x-26$

Therefore $f\left(x\right)=x^3-8x^2+25x-26$ have no negative real zero.

By synthetic division,

$f\left(x\right)=x^3-8x^2+25x-26$ in factored form $=\left(x^2-6x+13\right)\left(x-2\right)$.

The depressed equation $=\left(x^2-6x+13\right)$ .

$\left(x^2-6x+13\right)=$ quadratic equation $\left(a{x}^2+bx+c\right)$.

Here $a=1,b=-6,c=13$.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-\mathrm{4.1.13}}}{2.1}$

$\frac{-\left(-6\right)\pm \sqrt{36-52}}{2}=\frac{6\pm \sqrt{-16}}{2}$

The complex zeros $=3\pm 2i$.

Factored form of $f\left(x\right)=x^3-8x^2+25x-26$.

$=\left(x-2\right)\left[x-\left(3+2i\right)\right]\left[x-\left(3-2i\right)\right]=\left(x-2\right)\left[\left(x-3\right)-2i)\right]\left[\left(x-3\right)+2i)\right]$