Real Zeros and Multiplicity

Oct 13, 2025 by pinkandtrash.com 118 views

Introduction

In this article, we will delve into the analysis of a given polynomial function, $$f(x)=\frac{1}{3}\left(3 x^2+8\right)2\left(x^2+5\right)$$, to answer two crucial questions. Firstly, we will identify each real zero and its multiplicity. Secondly, we will determine whether the graph crosses or touches the $$x$$-axis at these zeros. This analysis will provide valuable insights into the behavior of the polynomial function and its graph.

Real Zeros and Multiplicity

To find the real zeros of the polynomial function, we need to set $$f(x)$$ equal to zero and solve for $$x$$. The given function can be rewritten as:

f(x)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=0

Since the function is a product of three factors, we can set each factor equal to zero to find the real zeros:

$\left(3 x^2+8\right)^2=0$
$\left(x^2+5\right)=0$

Solving the first equation, we get:

3 x^2+8=0

3 x^2=-8

x^2=-\frac{8}{3}

Since the square of any real number cannot be negative, there are no real solutions for this equation.

Solving the second equation, we get:

x^2+5=0

x^2=-5

Since the square of any real number cannot be negative, there are no real solutions for this equation.

However, we can rewrite the original function as:

f(x)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac<br/> # **Q&A: Analyzing the Polynomial Function** ## **Question 1: What are the real zeros of the polynomial function?** The real zeros of the polynomial function are the values of x that make the function equal to zero. In this case, we have: $$f(x)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=0

Since the function is a product of three factors, we can set each factor equal to zero to find the real zeros:

$\left(3 x^2+8\right)^2=0$
$\left(x^2+5\right)=0$

Solving the first equation, we get:

3 x^2+8=0

3 x^2=-8

x^2=-\frac{8}{3}

Since the square of any real number cannot be negative, there are no real solutions for this equation.

Solving the second equation, we get:

x^2+5=0

x^2=-5

Since the square of any real number cannot be negative, there are no real solutions for this equation.

However, we can rewrite the original function as:

f(x)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1}{3}\left(3 x^2+8\right)^2\left(x^2+5\right)=\frac{1