Solving Quartic Equation X^4 - 2x^3 - 19x^2 + 8x + 60 = 0

Introduction to Quartic Equations

A quartic equation, which is a polynomial equation of degree four, presents a unique challenge in the realm of algebra. Unlike quadratic equations, which have a straightforward formula for finding solutions, quartics often require more sophisticated techniques. This article delves into the intricacies of solving the specific quartic equation x4 - 2x3 - 19x2 + 8x* + 60 = 0*. We are given a crucial piece of information: two of the roots are α and -α, where α > 0. This condition significantly simplifies the problem and allows us to use algebraic manipulations and factorization to find all the roots. Understanding the nature of quartic equations is essential, as they appear in various fields of mathematics, physics, and engineering. The ability to solve them provides a powerful tool for tackling complex problems and gaining deeper insights into mathematical relationships. By exploring the step-by-step solution, we will not only find the roots of this specific equation but also gain a broader understanding of how to approach and solve similar quartic equations in the future.

Utilizing the Given Roots: α and -α

When tackling polynomial equations, identifying specific characteristics or relationships among the roots can significantly simplify the solution process. In our case, the quartic equation x4 - 2x3 - 19x2 + 8x* + 60 = 0* has a unique property: two of its roots are α and -α, where α is a positive real number. This information is invaluable because it tells us that the polynomial must have factors of the form (x - α) and (x + α). Multiplying these factors together, we obtain (x - α)(x + α) = x2 - α2*. This quadratic factor is a key component in breaking down the quartic equation. The presence of both a positive and negative root of the same magnitude indicates a symmetry within the equation, which can be exploited to reduce the complexity of the problem. By recognizing and utilizing this symmetry, we can transform the quartic equation into a more manageable form, ultimately leading to a straightforward solution. This approach highlights the importance of carefully examining the given information and identifying patterns that can aid in solving polynomial equations. The ability to recognize such patterns is a crucial skill in advanced algebra and problem-solving. Knowing that x2 - α2 is a factor allows us to perform polynomial division or factorization techniques to find the remaining factors and, consequently, the remaining roots of the equation.

Factorization and Polynomial Division

Having established that x2 - α2 is a factor of the quartic equation x4 - 2x3 - 19x2 + 8x* + 60 = 0*, our next step is to determine the value of α and find the other factor. To do this, we can utilize polynomial division or factorization techniques. Since α and -α are roots, when we substitute these values into the equation, the result should be zero. Let's denote the polynomial as P(x) = x4 - 2x3 - 19x2 + 8x + 60*. We know that P(α) = 0 and P(-α) = 0. However, directly substituting α into the equation might not immediately give us a clear value for α. Instead, we can use the fact that (x2 - α2)* is a factor to perform polynomial division. Dividing P(x) by (x2 - α2)* will yield a quadratic quotient, which we can then solve using standard methods. This process involves dividing the quartic polynomial by the quadratic factor to obtain another quadratic. The resulting quadratic equation can be solved using the quadratic formula or by factoring, giving us the remaining roots of the original quartic equation. This method of reducing a higher-degree polynomial into lower-degree polynomials is a fundamental technique in algebra. It allows us to systematically break down complex problems into simpler, more manageable parts. By performing polynomial division, we effectively reduce the quartic equation into a quadratic equation, which is much easier to solve. The quotient obtained from the division will be a quadratic expression, and finding its roots will give us the remaining solutions to the original quartic equation.

Finding α

To determine the value of α, we recognize that the constant term of the quartic polynomial is 60, and the constant term of the factor x2 - α2* is -α2. Therefore, the constant term of the quotient when we divide P(x) by x2 - α2* must be such that when multiplied by -α2, it equals 60. Let's denote the quotient as x2 + bx + c*. Then, (x2 - α2)(x2 + bx + c) = x4 - 2x3 - 19x2 + 8x + 60*. Expanding the left side, we get: x4 + bx3 + cx2 - α2x2 - βα2x - cα2 = x4 - 2x3 - 19x2 + 8x + 60*. Comparing the coefficients, we have the following equations:

• Coefficient of x3: b = -2
• Coefficient of x2: c - α2 = -19
• Coefficient of x: -βα2 = 8
• Constant term: -cα2 = 60

From the coefficient of x3, we find that b = -2. From the coefficient of x, we have -(-2)α2 = 8, which simplifies to 2α2 = 8. Dividing by 2, we get α2 = 4. Since α > 0, we find that α = 2. This crucial step allows us to specify the exact quadratic factor (x2 - 4), making the subsequent polynomial division precise and effective.

Performing Polynomial Division

Now that we have determined that α = 2, we know that x2 - 4* is a factor of the quartic polynomial. We can now perform polynomial division to find the other factor. Dividing x4 - 2x3 - 19x2 + 8x + 60* by x2 - 4*, we perform long division:

            x^2  - 2x  - 15
x^2 - 4 | x^4 - 2x^3 - 19x^2 + 8x + 60
          - (x^4         - 4x^2)
          -------------------------
                - 2x^3 - 15x^2 + 8x
                - (- 2x^3        + 8x)
                -------------------------
                      - 15x^2       + 60
                      - (- 15x^2       + 60)
                      -------------------------
                              0

The result of the division is x2 - 2x - 15*. This means that the original quartic equation can be written as (x2 - 4)(x2 - 2x - 15) = 0*. The polynomial division confirms that x2 - 4* is indeed a factor and provides us with the remaining quadratic factor, making it easier to find all the roots. This step is critical in reducing the complexity of the problem and setting the stage for finding the remaining solutions.

Solving the Quadratic Factors

After performing the polynomial division, we have successfully factored the original quartic equation into two quadratic factors: (x2 - 4)(x2 - 2x - 15) = 0*. To find the roots of the quartic equation, we now need to solve each quadratic factor separately. This approach simplifies the problem significantly, as solving quadratic equations is a well-established process. The first factor, x2 - 4*, is a simple difference of squares and can be easily factored as (x - 2)(x + 2) = 0. This gives us two roots: x = 2 and x = -2. These roots correspond to the given information that α and -α are roots, where α = 2. The second factor, x2 - 2x - 15*, is a standard quadratic equation that can be solved by factoring, completing the square, or using the quadratic formula. Factoring this quadratic involves finding two numbers that multiply to -15 and add to -2. These numbers are -5 and 3, so we can factor the quadratic as (x - 5)(x + 3) = 0. This gives us the remaining two roots: x = 5 and x = -3. By solving each quadratic factor, we have found all four roots of the original quartic equation. This step-by-step approach, from identifying the initial factors to solving the resulting quadratic equations, demonstrates a systematic method for tackling complex polynomial equations. The ability to break down a quartic equation into simpler quadratic factors is a crucial skill in algebra, allowing for a straightforward and accurate solution.

Determining All Roots of the Quartic Equation

By solving the quadratic factors, we have found all the roots of the quartic equation x4 - 2x3 - 19x2 + 8x* + 60 = 0*. The first quadratic factor, (x2 - 4), gave us the roots x = 2 and x = -2. These roots align with the given condition that two of the roots are α and -α, where α > 0. The second quadratic factor, x2 - 2x* - 15*, gave us the roots x = 5 and x = -3. Therefore, the complete set of roots for the quartic equation is x = -3, -2, 2, 5. This comprehensive solution demonstrates the effectiveness of using the given information about the roots to simplify the equation and systematically find all solutions. The combination of polynomial division, factorization, and solving quadratic equations provides a powerful toolkit for tackling quartic equations. Understanding how to apply these techniques allows us to approach complex algebraic problems with confidence and accuracy. The process of finding all roots not only solves the specific equation but also enhances our understanding of the fundamental principles of algebra and polynomial equations. This step-by-step solution highlights the importance of careful analysis, strategic application of algebraic techniques, and accurate execution to arrive at the final answer. The ability to find all roots of a polynomial equation is a cornerstone of algebraic problem-solving and is essential in various mathematical and scientific applications.

Conclusion: Significance of Solving Quartic Equations

In conclusion, solving the quartic equation x4 - 2x3 - 19x2 + 8x* + 60 = 0* using the given condition that two roots are α and -α (where α > 0) highlights the power of algebraic manipulation and factorization techniques. By recognizing the symmetry in the roots, we were able to simplify the equation and systematically find all solutions. The roots of the equation are x = -3, -2, 2, 5. This process demonstrates a comprehensive approach to solving quartic equations, which are an essential part of advanced algebra and have applications in various fields, including engineering, physics, and computer science. Understanding how to solve quartic equations provides a valuable skill set for tackling complex mathematical problems and gaining deeper insights into mathematical relationships. The ability to break down a complex equation into simpler parts, such as quadratic factors, is a fundamental technique in algebra. This method not only helps in solving the specific equation but also enhances our overall problem-solving abilities. The steps involved, from identifying factors to performing polynomial division and solving quadratic equations, illustrate a systematic approach that can be applied to a wide range of algebraic problems. Furthermore, the solution process underscores the importance of careful analysis, accurate execution, and strategic thinking in mathematics. The knowledge gained from solving quartic equations contributes to a broader understanding of polynomial equations and their applications in various scientific and technical disciplines. The ability to solve such equations is a testament to the power and elegance of algebraic methods.