Solving X^4 - 6x^3 - 3x^2 + 24x - 4 = 0 Finding Roots Step By Step

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Hey guys! Today, we're diving deep into the world of polynomial equations, specifically tackling the equation x^4 - 6x^3 - 3x^2 + 24x - 4 = 0. This might seem intimidating at first, but don't worry, we'll break it down step-by-step. Our mission is to find all the solutions (also known as roots) to this equation. We'll be using a combination of the Rational Root Theorem, synthetic division, and good old-fashioned algebraic manipulation. So, buckle up and let's get started!

a. Listing All Possible Rational Roots

When faced with a polynomial equation like this, the first thing we want to do is explore the possibility of rational roots. What exactly are rational roots, you ask? Well, they're roots that can be expressed as a fraction p/q, where p and q are integers. To help us find these potential roots, we turn to the Rational Root Theorem. This theorem is a real lifesaver, giving us a finite list of candidates to test.

The Rational Root Theorem states that if a polynomial equation with integer coefficients has a rational root p/q (in lowest terms), then p must be a factor of the constant term and q must be a factor of the leading coefficient. In our equation, x^4 - 6x^3 - 3x^2 + 24x - 4 = 0, the constant term is -4 and the leading coefficient is 1. This is a crucial starting point for solving polynomial equations.

So, let's identify the factors of -4. These are the possible values for 'p': ±1, ±2, ±4. Now, let's find the factors of the leading coefficient, which is 1. These are the possible values for 'q': ±1. According to the Rational Root Theorem, any rational root of our polynomial must be of the form p/q. To find all the possible rational roots, we take each factor of -4 and divide it by each factor of 1. Thus, the possible rational roots are ±1/1, ±2/1, ±4/1. Simplifying these fractions, we get the list of possible rational roots: ±1, ±2, ±4. Remember, these are just possibilities. We still need to test them to see if they actually are roots of the equation. Think of it like casting a wide net – we've identified the most likely candidates, and now we're going to check them out individually.

b. Using Synthetic Division to Test Possible Rational Roots

Now that we have our list of possible rational roots (±1, ±2, ±4), the next step is to test them. We need to figure out which, if any, of these numbers actually make the equation equal to zero. This is where synthetic division comes in handy. Synthetic division is a streamlined way to divide a polynomial by a linear factor (x - r), where 'r' is a potential root. It's much faster and less cumbersome than traditional long division, especially when dealing with higher-degree polynomials.

Let's start by testing the simplest possible root, 1. We set up the synthetic division as follows:

1 | 1  -6  -3  24  -4
  |      1  -5  -8  16
  ----------------------
    1  -5  -8  16  12

We bring down the leading coefficient (1), multiply it by the test root (1), and write the result (1) under the next coefficient (-6). We add these two numbers (-6 and 1) to get -5. We repeat the process: multiply -5 by 1 to get -5, write it under -3, add them to get -8, and so on. The last number in the bottom row (12) is the remainder. Since the remainder is not 0, 1 is not a root of the equation. Bummer!

Let's try -1 next:

-1 | 1  -6  -3  24  -4
   |     -1   7  -4 -20
   ----------------------
     1  -7   4  20 -24

Again, the remainder is -24, which is not 0, so -1 is not a root either. Don't get discouraged! This is a process of elimination.

Now, let's try 2:

2 | 1  -6  -3  24  -4
  |      2  -8 -22   4
  ----------------------
    1  -4 -11   2   0

Eureka! The remainder is 0. This means that 2 is a root of the equation. Not only that, but the bottom row of the synthetic division gives us the coefficients of the quotient polynomial. In this case, the quotient is x^3 - 4x^2 - 11x + 2. This is a crucial step because we've now reduced our fourth-degree polynomial to a third-degree polynomial, which is easier to work with. We have successfully identified an actual root using synthetic division!

c. Using the Root from Part (b) to Solve

We've discovered that 2 is a root of our equation, and we've also found the quotient polynomial, x^3 - 4x^2 - 11x + 2. This means we can rewrite our original equation as: (x - 2)(x^3 - 4x^2 - 11x + 2) = 0. To find the remaining roots, we need to solve the cubic equation x^3 - 4x^2 - 11x + 2 = 0. This is a significant simplification from our original fourth-degree equation.

Now, we can try to find another rational root for this cubic equation. We go back to the Rational Root Theorem. The factors of the constant term (2) are ±1, ±2, and the factors of the leading coefficient (1) are ±1. So, the possible rational roots are ±1, ±2. We already know 2 is a root of the original equation, so let's see if it's also a root of the cubic equation. We can use synthetic division again:

2 | 1  -4 -11   2
  |      2  -4 -30
  ----------------
    1  -2 -15 -28

The remainder is -28, so 2 is not a root of the cubic equation. Let's try -2:

-2 | 1  -4 -11   2
   |     -2  12  -2
   ----------------
     1  -6   1   0

Great! -2 is a root of the cubic equation. This means we can factor the cubic equation further. The quotient polynomial from this division is x^2 - 6x + 1. So, we can rewrite our equation as (x - 2)(x + 2)(x^2 - 6x + 1) = 0. We've now reduced the problem to solving a quadratic equation, x^2 - 6x + 1 = 0.

To solve the quadratic equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In our case, a = 1, b = -6, and c = 1. Plugging these values into the formula, we get:

x = (6 ± √((-6)^2 - 4 * 1 * 1)) / (2 * 1) x = (6 ± √(36 - 4)) / 2 x = (6 ± √32) / 2 x = (6 ± 4√2) / 2 x = 3 ± 2√2

So, the roots of the quadratic equation are 3 + 2√2 and 3 - 2√2. These are irrational roots.

Finally, we can list all the roots of the original equation, x^4 - 6x^3 - 3x^2 + 24x - 4 = 0:

  • x = 2
  • x = -2
  • x = 3 + 2√2
  • x = 3 - 2√2

And there you have it! We've successfully solved the fourth-degree polynomial equation by using the Rational Root Theorem, synthetic division, and the quadratic formula. It might have seemed daunting at first, but by breaking it down into smaller steps, we were able to find all the roots. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a polynomial-solving pro in no time!