**f(x)** = **(2x³ + 1 + x²)/(x² + 9) **as a sum of two **functions g(x) **and** h(x), **where g is an **odd function** and h is an **even function** is **f(x) = 2x³/(x² + 9) + (1 + x²)/(x² + 9).**

To answer the questions, we need to know what **functions** are

### What are functions?

A **function** is a mathematical expression that shows the relationship between two variables.

### Types of function

**Odd functions**- These are functions in which f(-x) = -f(x)**Even functions -**These are functions in which f(-x) = f(x)**Neither -**When the function is neither even nor odd.

Since **f(x) = (2x³ + 1 + x²)/(x² + 9).**

### The odd part of f(x)

Its odd part **g(x) = [f(x) - f(-x)]/2**

**So, g(x) = [f(x) - f(-x)]/2 **

= [(2x³ + 1 + x²)/(x² + 9) - (2(-x)³ + 1 + (-x)²)/((-x)² + 9)]/2

= [(2x³ + 1 + x²)/(x² + 9) - (-2x³ + 1 + x²)/(x² + 9)]/2

= (2x³ + 1 + x² + 2x³ - 1 - x²)/2(x² + 9)

= 4x³/2(x² + 9)

**g(x) = 2x³/(x² + 9) **

**The even part of f(x)**

Its even part **h(x) = [f(x) + f(-x)]/2**

**So, h(x) = [f(x) + f(-x)]/2 **

= [(2x³ + 1 + x²)/(x² + 9) + (2(-x)³ + 1 + (-x)²)/((-x)² + 9)]/2

= [(2x³ + 1 + x²)/(x² + 9) + (-2x³ + 1 + x²)/(x² + 9)]/2

= (2x³ + 1 + x² - 2x³ + 1 + x²)/2(x² + 9)

= 2(1 + x²)/2(x² + 9)

**h(x) = (1 + x²)/(x² + 9) **

So,** f(x) = g(x) + h(x)**

**f(x) = 2x³/(x² + 9) + (1 + x²)/(x² + 9) **

So, **f(x)** = **(2x³ + 1 + x²)/(x² + 9) **as a sum of two **functions g(x) **and** h(x), **where g is an **odd function** and h is an **even function** is **f(x) = 2x³/(x² + 9) + (1 + x²)/(x² + 9).**

Learn more about **odd** and **even functions** here:

https://brainacademy.pro/question/26322830

Use longhand division or synthetic division. Should help simply.