Chapter 4.3, Problem 25AYU

To determine

To find: The remaining zeros of the polynomial function $f\left(x\right)$ whose coefficients are real numbers.

Answer to Problem 25AYU

$\text{Factored form of }f\left(x\right)=2x^4+5x^3+5x^2+20x-12=\left(x+3\right)\left(x-2i\right)\left(x+2i\right)\left(2x-1\right)$.

Explanation of Solution

Given:

$f\left(x\right)=2x^4+5x^3+5x^2+20x-12$

$\text{Zero }=-2i$.

Degree of the polynomial function $f\left(x\right)=2x^4+5x^3+5x^2+20x-12=4$.

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

From Descartes’ rule of signs.

$f\left(x\right)=2x^4+5x^3+5x^2+20x-12$

$+\text{ To }–$

Therefore $f\left(x\right)=2x^4+5x^3+5x^2+20x-12$ have 1 positive real zero.

$f\left(-x\right)=2x^4-5x^3+5x^2-20x-12$

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

Therefore $f\left(x\right)=2x^4+5x^3+5x^2+20x-12$ have 3 or 1 negative real zeros.

Rational zeros theorem provides information about the potential rational zeros of the polynomial function with integer coefficients.

Zeros of $a_0=p=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$.

Zeros of $a_n=q=\pm 1,\pm 2$.

$\frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12,\pm \frac{1}{2}\pm \frac{3}{2}$

By Synthetic division.

The coefficients of $f\left(x\right)=25520-12$.

The negative real zero $=-3$.

The depressed equation $=2x^3-x^2+8x-4$.

The complex zero $=-2i$.

From conjugate pair theorem the complex zero occur only in complex pairs.

The complex zeros $=\pm 2i$.

The real zeros $=-3$.

Factored form of $f\left(x\right)=2x^4+5x^3+5x^2+20x-12=\left(x+3\right)\left(x-2i\right)\left(x+2i\right)\left(2x-1\right)$.