## One root of f(x) = 2×3 + 9×2 + 7x – 6 is –3. explain how to find the factors of the polynomial

Polynomials serve as the building blocks of algebra, appearing in various mathematical contexts and real-world applications. Among the myriad of polynomial equations, understanding how to find their roots—the values of

�

*x* that make the polynomial equal to zero—remains a fundamental skill. In this guide, we will delve into the process of demystifying polynomial factors, focusing on the polynomial equation

�(�)=2�3+9�2+7�–6

*f*(*x*)=2*x*

3

+9*x*

2

+7*x*–6.

**Understanding Polynomial Roots**

Before we embark on our journey to uncover the roots of the polynomial, let’s establish a clear understanding of what polynomial roots entail. A root of a polynomial is a value of

�

*x* that, when substituted into the polynomial equation, yields an output of zero. In essence, finding the roots involves solving the equation

�(�)=0

*f*(*x*)=0, where

�(�)

*f*(*x*) represents the polynomial function.

**Exploring the Polynomial:**

### �(�)=2�3+9�2+7�–6

*f*(*x*)=2*x*

### 3

### +9*x*

### 2

### +7*x*–6

Our focus lies in unraveling the roots of the polynomial

�(�)=2�3+9�2+7�–6

*f*(*x*)=2*x*

3

+9*x*

2

+7*x*–6. To begin our exploration, let’s break down the components of this polynomial equation. Here, the highest exponent of

�

*x* is 3, indicating that this is a cubic polynomial.

**Strategies for Finding Roots**

There exist several methods for finding polynomial roots, each with its advantages and applications. Some common techniques include factoring, synthetic division, graphical analysis, and numerical methods such as Newton’s method or the bisection method. In this guide, we’ll focus on factoring and synthetic division.

**Factoring the Polynomial**

One approach to finding polynomial roots involves factoring the polynomial expression. By factoring, we aim to break down the polynomial into simpler expressions, revealing its roots more explicitly. Let’s explore how we can factor the polynomial

2�3+9�2+7�–6

2*x*

3

+9*x*

2

+7*x*–6 to uncover its roots.

**Synthetic Division: A Tool for Dividing Polynomials**

Another powerful tool in our arsenal for finding polynomial roots is synthetic division. Synthetic division allows us to divide a polynomial by a linear expression, often simplifying the process of factoring and identifying roots. We’ll utilize synthetic division to further explore the roots of

�(�)=2�3+9�2+7�–6

*f*(*x*)=2*x*

3

+9*x*

2

+7*x*–6.

**Step-by-Step Approach**

Let’s outline a step-by-step approach to finding the roots of the polynomial

�(�)=2�3+9�2+7�–6

*f*(*x*)=2*x*

3

+9*x*

2

+7*x*–6:

- Substitute
- �=0
*x*=0: Begin by substituting- �=0
*x*=0 into the polynomial equation to determine if zero is a root.- Apply Factoring Techniques: Utilize factoring techniques such as grouping, difference of squares, or trinomial factoring to break down the polynomial expression.
- Use Synthetic Division: If factoring seems challenging, resort to synthetic division to divide the polynomial by potential roots or factors.
- Iterative Methods: Employ iterative numerical methods or graphical analysis to approximate roots if analytical methods prove difficult or impractical.
- Verify Solutions: Once potential roots are identified, verify them by substituting back into the original polynomial equation to ensure they yield zero.

**Case Study: Finding Roots of**

### �(�)=2�3+9�2+7�–6

*f*(*x*)=2*x*

### 3

### +9*x*

### 2

### +7*x*–6

Let’s apply the strategies outlined above to find the roots of the polynomial

�(�)=2�3+9�2+7�–6

*f*(*x*)=2*x*

3

+9*x*

2

+7*x*–6.

#### Step 1: Substituting

#### �=0

*x*=0

Substituting

�=0

*x*=0 into the polynomial equation yields:

�(0)=2(0)3+9(0)2+7(0)–6

*f*(0)=2(0)

3

+9(0)

2

+7(0)–6

�(0)=−6

*f*(0)=−6

Since

�(0)≠0

*f*(0)

=0,

�=0

*x*=0 is not a root of the polynomial.

#### Step 2: Factoring and Synthetic Division

Factoring the polynomial

2�3+9�2+7�–6

2*x*

3

+9*x*

2

+7*x*–6 directly might be challenging due to its cubic nature. Instead, we can employ synthetic division to test potential roots.

#### Synthetic Division with Potential Root

#### �=−3

*x*=−3

Using synthetic division with

�=−3

*x*=−3 as a potential root:

-3 | 2 9 7 -6

|__________

| 0 -6 -48

The result suggests that

�=−3

*x*=−3 is a root of the polynomial, leading to a quotient of

2�2−6�−48

2*x*

2

−6*x*−48.

#### Solving the Quadratic Equation

Factoring the quadratic expression

2�2−6�−48=0

2*x*

2

−6*x*−48=0, we obtain:

2(�−6)(�+4)=0

2(*x*−6)(*x*+4)=0

This yields the roots

�=6

*x*=6 and

�=−4

*x*=−4.

**Conclusion: Unlocking Polynomial Roots**

In this guide, we’ve embarked on a journey to demystify polynomial factors and uncover the roots of the polynomial

�(�)=2�3+9�2+7�–6

*f*(*x*)=2*x*

3

+9*x*

2

+7*x*–6. Through a systematic approach involving factoring, synthetic division, and analytical techniques, we’ve elucidated the process of finding polynomial roots and showcased the significance of these roots in understanding polynomial behavior. Armed with these insights and strategies, you’re now equipped to tackle polynomial equations with confidence, unlocking the mysteries of polynomial factors along the way.