Chapter 4.3, Problem 24AYU

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

$g\left(x\right)=x^3+3x^2+25x+75$; zero: $-5i$

To determine

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

Answer to Problem 24AYU

$\text{Factored form of }g(x)=x^3+3x^2+25x+75=(x+3)(x-5i)(x+5i)$.

Explanation of Solution

Given:

$g\left(x\right)=x^3+3x^2+25x+75$

$\text{Zero }=-5i$.

Degree of the polynomial function $g\left(x\right)=x^3+3x^2+25x+75=3$.

Therefore $g\left(x\right)$ have 3 zeros.

From Descartes’ rule of signs.

$g\left(-x\right)=-x^3+3x^2-25x+75$

$-To++to–$

Therefore $g\left(x\right)=x^3+3x^2+25x+75$ have 2 negative real zeros or no negative real zeros.

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

Zeros of $a_0=p=\pm 1,\pm 3,\pm 5,\pm 15,\pm 25,\pm 75$.

Zeros of an $a_n=q=\pm 1$.

By Synthetic division.

The coefficients of $g\left(x\right)=132575$.

The negative real zero $=-3$.

The depressed equation $=x^2+25$.

The depressed equation can be factored further $=x^2=-25x=\pm 5i$.

The complex zeros $=\pm 5i$.

The real zeros $=-3$.

Factored form of $g\left(x\right)=x^3+3x^2+25x+75=\left(x+3\right)\left(x–5i\right)\left(x+5i\right)$.