Chapter 4.3, Problem 39AYU

To determine

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

Answer to Problem 39AYU

$\text{The complex zeros = 2 ± 3i}$,

$\text{Factored form of f}\left(x\right) = 3x^4 - x^3 - 9x^2 + 159x - 52$

$\left[\left(x - 2\right) - 3i)\right] [\left(x - 2\right) + 3i]\left[x - \frac{1}{3}\right]\left[\text{x} + 4\right]$

Explanation of Solution

Given:

$f\left(x\right) = 3x^4 - x^3 - 9x^2 + 159x - 52$

Degree of the polynomial function $f\left(x\right) = 3x^4 - x^3 - 9x^2 + 159x - 52 = 4$.

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

From Descartes’ rule of signs

$f\left(x\right) = 3x^4 - x^3 - 9x^2 + 159x - 52$

$+\text{ To }- -\text{ to }+ +\text{ to }-$

Therefore $f\left(x\right) = 3x^4 - x^3 - 9x^2 + 159x - 52$ have 3 or 1 positive real zeros

$f\left(-x\right) = 3x^4 + x^3 - 9x^2 - 159x - 52$

$+\text{ TO }-$

Therefore $f\left(x\right) = 3x^4 - x^3 - 9x^2 + 159x - 52$ have 1 negative real zero

By synthetic division

The real positive zero $=\mathrm{-4}$

The depressed equation $3x^3 - 13x^2 + 43x - 13$

The real negative zero $= \frac{1}{3}$

Depressed equation $= 3x^2 - 12x + 39 = 3\left(x^2 - 4x + 13\right)$

Quadratic equation $a{x}^2 + bx + c$

$\text{a = 1, b = }-4,\text{ c = }13$

$x = \frac{-b \pm \sqrt{b^2 - \mathrm{4 . }a . c}}{\mathrm{2 . }a} = \frac{-\left(-4\right) \pm \sqrt{{\left(-4\right)}^2 - \mathrm{4 . 1 . 13}}}{\mathrm{2 . 1}} = \frac{4 \pm \sqrt{-36}}{2}$

The complex zeros $\text{= 2 ± 3i}$,

Factored form of $\text{f}\left(x\right) = 3x^4 - x^3 - 9x^2 + 159x - 52$

$= [x - \left(2 + 3i\right)] \left[x - \left(2 - 3i\right)\right]\left[x - \frac{1}{3}\right]\left[x - \left(-4\right)\right]$

$= \left[\left(x - 2\right) - 3i)\right]\left[\left(x - 2\right) + 3i\right]\left[x - \frac{1}{3}\right]\left[x + 4\right]$