Chapter 4.3, Problem 40AYU

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 40AYU

$\text{The complex zeros = 2 ± 2i}$,

$\text{Factored form of f}\left(x\right) = 2x^4 + x^3 - 35x^2 - 113x + 65$

$= \left[\left(x - 2\right) - 2i)\right]\left[\left(x - 2\right) + 2i\right]\left[x - \frac{1}{2}\right]\left[x - 5\right]$

Explanation of Solution

Given:

$f\left(x\right) = 2x^4 + x^3 - 35x^2 - 113x + 65$

Degree of the polynomial function $f\left(x\right) = 2x^4 + x^3 - 35x^2 - 113x + 65 = 4$.

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

From Descartes’ rule of signs.

$f\left(x\right) = 2x^4 + x^3 - 35x^2 - 113x + 65$

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

Therefore $f\left(x\right) = 2x^4 + x^3 - 35x^2 - 113x + 65$ have 2 or 0 positive real zeros.

$f\left(-x\right) = 2x^4 - x^3 - 35x^2 + 113x + 65$

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

Therefore $f\left(x\right) = 2x^4 + x^3 - 35x^2 - 113x + 65$ have 2 or 0 negative real zero.

By synthetic division.

The real positive zero $=5$.

The depressed equation $= 2x^3 + 11x^2 + 20x - 13$.

The real negative zero $= \frac{1}{2}$.

Depressed equation $= 2x^2 + 12x + 26 = 2\left(x^2 + 6x + 13\right)$.

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

$\text{a = 1, b = }-6,\text{ c = }13$

$x = \frac{-b \pm \sqrt{b^2 - \mathrm{4 . }a . c}}{\mathrm{2 . }a} = \frac{-\left(6\right) \pm \sqrt{{\left(6\right)}^2 - \mathrm{4 . 1 . 13}}}{\mathrm{2 . 1}} = \frac{4 \pm \sqrt{-16}}{2}$

The complex zeros $\text{= 2 ± 2i}$,

Factored form of $\text{f}\left(x\right) = 2x^4 + x^3 - 35x^2 - 113x + 65$.

$= \left[x - \left(2 + 2i\right)\right]\left[x - \left(2 - 2i\right)\right]\left[x - \frac{1}{2}\right]\left[x - \left(5\right)\right]$

$= \left[\left(x - 2\right) - 2i)\right]\left[\left(x - 2\right) + 2i\right]\left[x\hspace{0.17em}-\hspace{0.17em}\frac{1}{2}\right]\left[x\hspace{0.17em}-\hspace{0.17em}5\right]$