Finding The Other Roots Of 2x³ - 9x² + 16x - 12 = 0

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In the realm of mathematics, solving polynomial equations is a fundamental skill. Cubic equations, in particular, present a unique challenge due to their degree. This article delves into the process of finding all roots of the cubic equation 2x³ - 9x² + 16x - 12 = 0, given that one root is already known. We'll explore the techniques of polynomial division and the quadratic formula to unveil the remaining roots, offering a comprehensive guide for students and math enthusiasts alike.

Understanding Cubic Equations

Cubic equations, which are polynomial equations of the third degree, take the general form ax³ + bx² + cx + d = 0, where a ≠ 0. Solving these equations involves finding the values of x that satisfy the equation, known as the roots or solutions. A cubic equation can have up to three roots, which may be real or complex.

Finding the roots of a cubic equation can be more complex than solving quadratic equations. However, if one root is known, we can reduce the cubic equation to a quadratic equation, which can then be solved using the quadratic formula or factoring. This approach simplifies the problem and allows us to find all the roots systematically.

Step 1: Polynomial Division

Polynomial division is the cornerstone of our solution. We're given that x = 2 is a root of the equation 2x³ - 9x² + 16x - 12 = 0. This implies that (x - 2) is a factor of the polynomial. We'll use polynomial long division (or synthetic division) to divide the cubic polynomial by (x - 2).

Performing Polynomial Long Division

To divide 2x³ - 9x² + 16x - 12 by (x - 2), we set up the long division as follows:

 x - 2 | 2x³ - 9x² + 16x - 12

  1. Divide the leading term: Divide 2x³ by x, which gives 2x². Write this above the -9x² term.

2x² x - 2 | 2x³ - 9x² + 16x - 12 ```

  1. Multiply and subtract: Multiply (x - 2) by 2x², resulting in 2x³ - 4x². Subtract this from the original polynomial.

2x² x - 2 | 2x³ - 9x² + 16x - 12 - (2x³ - 4x²) ----------------- -5x² + 16x ```

  1. Bring down the next term: Bring down the +16x term.

2x² x - 2 | 2x³ - 9x² + 16x - 12 - (2x³ - 4x²) ----------------- -5x² + 16x ```

  1. Repeat the process: Divide -5x² by x, which gives -5x. Write this next to the 2x² term.

2x² - 5x x - 2 | 2x³ - 9x² + 16x - 12 - (2x³ - 4x²) ----------------- -5x² + 16x ```

  1. Multiply and subtract: Multiply (x - 2) by -5x, resulting in -5x² + 10x. Subtract this from -5x² + 16x.

2x² - 5x x - 2 | 2x³ - 9x² + 16x - 12 - (2x³ - 4x²) ----------------- -5x² + 16x - (-5x² + 10x) ----------------- 6x - 12 ```

  1. Bring down the last term: Bring down the -12 term.

2x² - 5x x - 2 | 2x³ - 9x² + 16x - 12 - (2x³ - 4x²) ----------------- -5x² + 16x - (-5x² + 10x) ----------------- 6x - 12 ```

  1. Repeat again: Divide 6x by x, which gives 6. Write this next to the -5x term.

2x² - 5x + 6 x - 2 | 2x³ - 9x² + 16x - 12 - (2x³ - 4x²) ----------------- -5x² + 16x - (-5x² + 10x) ----------------- 6x - 12 ```

  1. Multiply and subtract: Multiply (x - 2) by 6, resulting in 6x - 12. Subtract this from 6x - 12.

2x² - 5x + 6 x - 2 | 2x³ - 9x² + 16x - 12 - (2x³ - 4x²) ----------------- -5x² + 16x - (-5x² + 10x) ----------------- 6x - 12 - (6x - 12) ----------------- 0 ```

The remainder is 0, which confirms that (x - 2) is indeed a factor. The quotient is 2x² - 5x + 6. This means we can rewrite the original equation as:

(x - 2)(2x² - 5x + 6) = 0

Step 2: Solving the Quadratic Equation

Solving the quadratic equation is the next crucial step. We now have a quadratic equation, 2x² - 5x + 6 = 0. To find the remaining roots, we can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 2, b = -5, and c = 6. Plugging these values into the quadratic formula, we get:

x = (5 ± √((-5)² - 4 * 2 * 6)) / (2 * 2) x = (5 ± √(25 - 48)) / 4 x = (5 ± √(-23)) / 4

Since the discriminant (the value inside the square root) is negative, we have complex roots. We can rewrite √(-23) as i√23, where i is the imaginary unit (√-1).

Therefore, the roots are:

x = (5 ± i√23) / 4 x = 5/4 ± (i√23) / 4

Step 3: Identifying All Roots

Identifying all roots requires us to combine the given root with the ones we have calculated. We've found two complex roots and were given one real root. So, the roots of the equation 2x³ - 9x² + 16x - 12 = 0 are:

  1. x = 2
  2. x = 5/4 + (i√23) / 4
  3. x = 5/4 - (i√23) / 4

Conclusion

In conclusion, by utilizing polynomial division and the quadratic formula, we have successfully determined all three roots of the cubic equation 2x³ - 9x² + 16x - 12 = 0. The roots are 2, 5/4 + (i√23) / 4, and 5/4 - (i√23) / 4. This process demonstrates a systematic approach to solving cubic equations when one root is known, highlighting the interplay between algebraic techniques and complex numbers. Understanding these methods is crucial for advanced mathematical problem-solving and provides a solid foundation for further studies in algebra and calculus.

Frequently Asked Questions (FAQ)

Q1: What is a cubic equation?

A cubic equation is a polynomial equation of the third degree. Its general form is ax³ + bx² + cx + d = 0, where a ≠ 0. Cubic equations can have up to three roots, which may be real or complex numbers.

Q2: How many roots can a cubic equation have?

A cubic equation can have up to three roots. These roots can be real, complex, or a combination of both. Complex roots always occur in conjugate pairs if the coefficients of the equation are real.

Q3: What is polynomial division and why is it used in solving cubic equations?

Polynomial division is a method used to divide one polynomial by another. In the context of solving cubic equations, if one root is known, polynomial division is used to divide the cubic polynomial by a linear factor (x - root), reducing it to a quadratic equation. This simplifies the process of finding the remaining roots.

Q4: Can synthetic division be used instead of polynomial long division?

Yes, synthetic division can be used as an alternative to polynomial long division, especially when dividing by a linear factor of the form (x - k). Synthetic division is a more streamlined and quicker method for this specific case.

Q5: What is the quadratic formula and when should it be used?

The quadratic formula is a formula used to find the roots of a quadratic equation of the form ax² + bx + c = 0. The formula is x = (-b ± √(b² - 4ac)) / (2a). It should be used when factoring the quadratic equation is difficult or not possible, and it guarantees finding the roots, whether they are real or complex.

Q6: What is the discriminant and what does it tell us about the roots of a quadratic equation?

The discriminant is the expression inside the square root in the quadratic formula (b² - 4ac). It provides information about the nature of the roots:

  • If the discriminant is positive, the quadratic equation has two distinct real roots.
  • If the discriminant is zero, the quadratic equation has one real root (a repeated root).
  • If the discriminant is negative, the quadratic equation has two complex conjugate roots.

Q7: What are complex roots and how do they arise in cubic equations?

Complex roots are roots that include an imaginary part, represented by the imaginary unit i, where i = √-1. Complex roots arise in cubic equations when the discriminant of the quadratic equation (obtained after polynomial division) is negative. Complex roots always occur in conjugate pairs if the coefficients of the polynomial are real.

Q8: How do you handle complex roots when solving a cubic equation?

When complex roots arise, you continue to apply the quadratic formula, which will yield roots in the form a + bi and a - bi, where a and b are real numbers and i is the imaginary unit. These complex roots are valid solutions to the equation.

Q9: Is it possible for a cubic equation to have only one real root?

Yes, it is possible for a cubic equation to have only one real root. This occurs when the other two roots are complex conjugates. For instance, if the cubic equation has real coefficients, any complex roots must occur in conjugate pairs, meaning there will always be an even number of complex roots.

Q10: What are the key steps to solve a cubic equation if one root is known?

The key steps to solve a cubic equation if one root is known are:

  1. Use the known root to form a linear factor (x - root).
  2. Perform polynomial division (long division or synthetic division) to divide the cubic polynomial by the linear factor, resulting in a quadratic quotient.
  3. Solve the quadratic equation using the quadratic formula or factoring.
  4. Combine the known root with the roots obtained from the quadratic equation to list all roots of the cubic equation.

By following these FAQs, readers can gain a deeper understanding of cubic equations and the methods used to solve them, including polynomial division, the quadratic formula, and the nature of complex roots. This knowledge is essential for students and anyone interested in advanced mathematical problem-solving.