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

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

Understanding and solving polynomials is a fundamental aspect of algebra, and one intriguing polynomial that often appears in mathematical studies is f(x) = 2x^3 + 9x^2 + 7x – 6. In this exploration, we will delve into the process of solving for the root when x equals -3, unraveling the intricate web of factors that constitute the polynomial.

 

The Foundation: Polynomial Basics

Before we embark on solving the equation, let’s establish a solid foundation by revisiting some polynomial basics. A polynomial is a mathematical expression consisting of variables, coefficients, and exponents, and it often plays a pivotal role in various mathematical models and real-world scenarios. The power of the polynomial, represented by the highest exponent, influences its behavior and the number of roots it possesses.

Breaking Down the Polynomial: f(x) = 2x^3 + 9x^2 + 7x – 6

Our focus lies in the polynomial f(x) = 2x^3 + 9x^2 + 7x – 6, and our goal is to find the root when x is set to -3. To begin, we need to understand the anatomy of the polynomial. The highest exponent here is 3, indicating that this is a cubic polynomial. Cubic polynomials can have up to three roots, and our task is to pinpoint one of them when x is -3.

Expressing the Equation: f(-3) = 2(-3)^3 + 9(-3)^2 + 7(-3) – 6

To find the root at x = -3, substitute -3 into the equation for x:

�(−3)=2(−3)3+9(−3)2+7(−3)–6

f(−3)=2(−3)

3

+9(−3)

2

+7(−3)–6

This expression may seem complex, but fear not; we will break it down step by step.

Step 1: Compute the Powers

Firstly, calculate the powers of -3:

(−3)3=−27

(−3)

3

=−27

(−3)2=9

(−3)

2

=9

Step 2: Substitute and Simplify

Now, substitute these values back into the original equation:

�(−3)=2(−27)+9(9)+7(−3)–6

f(−3)=2(−27)+9(9)+7(−3)–6

Step 3: Simplify Further

Continue simplifying:

�(−3)=−54+81−21–6

f(−3)=−54+81−21–6

�(−3)=0

f(−3)=0

Surprisingly, the result is zero! This implies that -3 is indeed a root of the polynomial. But the journey doesn’t end here; we must delve deeper to understand the factors that lead to this outcome.

Factorizing the Polynomial

To comprehend how -3 became a root, we must factorize the polynomial. Factorization involves breaking down a polynomial into its constituent factors, revealing the roots more explicitly.

Synthetic Division: Unveiling the Factors

One method to factorize polynomials efficiently is synthetic division. Using synthetic division with -3 as the root, we can uncover the quotient and remainders that expose the factors.

2�3+9�2+7�–6 divided by (�+3)

2x

3

+9x

2

+7x–6divided by(x+3)

Let’s perform synthetic division:

-3 | 2    9    7   -6

|__________

|  0  -6   -48

The result indicates that the quotient is

2�2−6�−48

2x

2

−6x−48. Now, this quadratic expression represents the factorized form of the original polynomial.

Solving the Quadratic Equation

To find the remaining factors, we need to solve the quadratic equation

2�2−6�−48=0

2x

2

−6x−48=0. This equation can be factored into

(�−6)(2�+8)=0

(x−6)(2x+8)=0, yielding the factors

(�−6)

(x−6) and

(2�+8)

(2x+8).

The Final Insight: Understanding the Roots

Combining all factors, the original polynomial

2�3+9�2+7�–6

2x

3

+9x

2

+7x–6 can be expressed as

(�+3)(�−6)(2�+8)

(x+3)(x−6)(2x+8). Now, it becomes evident why -3 is a root; when

(�+3)

(x+3) is set to zero,

x equals -3, confirming our initial observation.

 

Conclusion

In this journey of unraveling polynomial roots, we’ve navigated through the intricacies of

�(�)=2�3+9�2+7�–6

f(x)=2x

3

+9x

2

+7x–6, specifically when x is -3. From the initial substitution to the synthetic division and factoring, each step has contributed to our understanding of how -3 became a root and how the polynomial can be expressed in its factored form. This exploration not only enhances our grasp of algebraic concepts but also provides a roadmap for tackling similar polynomial challenges in the future.

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