Chapter 4.3, Problem 30AYU

In Problems 23-30, use the given zero to find the remaining zeros of each function.

$g\left(x\right)=2x^5-3x^4-5x^3-15x^2-207x+108$; zero: $3i$

To determine

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

Answer to Problem 30AYU

$\text{Factored form of }g\left(x\right)=2x^5-3x^4-5x^3-15x^2-207x+108$.

$=\left(x+3\right)\left(x^2+9\right)\left(x-4\right)\left(2x-1\right)$

Explanation of Solution

Given:

$g\left(x\right)=2x^5-3x^4-5x^3-15x^2-207x+108$

$\text{Zero }=3i$.

Degree of the polynomial function $g\left(x\right)=2x^5-3x^4-5x^3-15x^2-207x+108=5$.

Therefore $g\left(x\right)$ have 5 zeros.

From Descartes’ rule of signs.

$g\left(x\right)=2x^5-3x^4-5x^3-15x^2-207x+108$

$+To–-to+$

Therefore $g\left(x\right)=2x^5-3x^4-5x^3-15x^2-207x+108$have 2 or 0 positive real zero.

$g\left(-x\right)=-2x^5-3x^4+5x^3-15x^2+207x+108$

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

Therefore $g\left(x\right)=2x^5-3x^4-5x^3-15x^2-207x+108$ 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 $a_0=p=\pm 1,\pm 2,\pm 3,\pm 6,\pm 9\pm 12,\pm 18,\pm 27,\pm 36,\pm 54,\pm 108$.

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

$\frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 12,\pm 18,\pm 27,\pm 36,\pm 54,\pm 108,\pm \frac{1}{2}\pm \frac{3}{2},\pm \frac{9}{2}$

By Synthetic division.

The coefficients of $g\left(x\right)=2-3-5-15-207108$.

The positive real zero $=4$.

The depressed equation $=2x^4+5x^3+15x^2+45x-27$.

The negative real zero $=-3$.

The depressed equation $=2x^3-x^2+18-9$.

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

The complex zeros $=3i,-3i$.

Factored form of $g\left(x\right)=2x^5-3x^4-5x^3-15x^2-207x+108$.

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

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

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

$i^2= -1$

$g\left(x\right)=\left(x+3\right)\left(x^2+9\right)\left(x-4\right)\left(2x-1\right)$