Chapter 4.3, Problem 29AYU

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

$h\left(x\right)=3x^5+2x^4+15x^3+10x^2-528x-352$; zero: $-4i$

To determine

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

Answer to Problem 29AYU

$\text{Factored form of }h\left(x\right)=3x^5+2x^4+15x^3+10x^2-528x-352$.

$=\left(3x+2\right)\left[x-\left(-4i\right)\right]\left[x-4i\right]\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)$

Explanation of Solution

Given:

$h\left(x\right)=3x^5+2x^4+15x^3+10x^2-528x-352$

$\text{Zero }=-4i$.

Degree of the polynomial function $h\left(x\right)=3x^5+2x^4+15x^3+10x^2-528x-352=5$.

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

From Descartes’ rule of signs.

$h\left(x\right)=3x^5+2x^4+15x^3+10x^2-528x-352$

$+\text{ To }–$

Therefore $h\left(x\right)=3x^5+2x^4+15x^3+10x^2-528x-352$ have 1 positive real zero.

$h\left(-x\right)=-3x^5+2x^4-15x^3+10x^2+528x-352$

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

Therefore $h\left(x\right)=3x^5+2x^4+15x^3+10x^2-528x-352$ have 4 or 2 negative real zeros.

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

Zeros of $a_0=p=\pm 1,\pm 2,\pm 4,\pm 8,\pm 11\pm 16,\pm 22,\pm 32,\pm 44,\pm 88,\pm 176,\pm 352$.

Zeros of $a_n=q=\pm 1,\pm 3$.

$\frac{p}{q}=\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3 },\pm \frac{8}{3},\pm \frac{11}{3}\pm 1,\pm 2,\pm 4,\pm 8,\pm 11\pm 16,\pm 22,\pm 32,\pm 44,\pm 88,\pm 176,\pm 352\dots \dots \mathrm{..}$

By Synthetic division.

The coefficients of $h\left(x\right)=03+21510-528-352$.

The complex zero $=-4i$.

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

The complex zeros $=-4i,+4i$.

The depressed equation $=3x^3+2x^2-33x-22$.

Factored form of

$h\left(x\right)=3x^5+2x^4+15x^3+10x^2-528x-352$

$=\left(3x+2\right)\left[x-\left(-4i\right)\right]\left[x-4i\right]\left(x-\sqrt{11}\right)\left(x+\sqrt{11}\right)$