Solve: $x^4+6x^3-3x^2-24x-4=0$ | Rational Root Theorem

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Hey guys! Today, we're diving into the fascinating world of polynomial equations, specifically this quartic equation: $x^4 + 6x^3 - 3x^2 - 24x - 4 = 0$. Now, I know what you might be thinking: quartic equations can look super intimidating, but don't worry! We're going to break it down step by step using the Rational Root Theorem and some clever synthetic division. So, grab your pencils, and let's get started!

a) Listing Possible Rational Roots

First things first, we need to figure out what our potential rational roots are. This is where the Rational Root Theorem comes to the rescue. This theorem is a fantastic tool that helps us narrow down the possibilities. Basically, it tells us that any rational root of the polynomial equation (if it exists) must be of the form p/q, where 'p' is a factor of the constant term and 'q' is 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. So, let's list the factors:

• Factors of the constant term (-4): ±1, ±2, ±4
• Factors of the leading coefficient (1): ±1

Now, we form all possible fractions p/q by dividing each factor of -4 by each factor of 1. This gives us our list of possible rational roots:

Possible Rational Roots: ±1, ±2, ±4

That's it! We've narrowed down our search to just these six numbers. It's way better than randomly guessing, right? But remember, these are just possible roots. We need to test them to see if they actually work.

b) Using Synthetic Division to Identify an Actual Root

Alright, now comes the fun part: testing our possible roots. We're going to use a technique called synthetic division, which is a neat and efficient way to divide a polynomial by a linear factor (x - r), where 'r' is our potential root. If the remainder after the division is zero, then 'r' is indeed a root of the equation. Let's dive right in! We'll start by testing the simplest root first, which is 1.

Testing x = 1

Set up the synthetic division table with the coefficients of our polynomial (1, 6, -3, -24, -4) and the potential root (1):

1 | 1  6  -3  -24  -4
  |        
  ----------------------

Bring down the first coefficient (1):

1 | 1  6  -3  -24  -4
  |        
  ----------------------
  1

Multiply the root (1) by the number we just brought down (1), and write the result (1) under the next coefficient (6):

1 | 1  6  -3  -24  -4
  |  1      
  ----------------------
  1

Add the numbers in the second column (6 + 1 = 7):

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

Repeat the process: multiply the root (1) by the result (7), write it under the next coefficient (-3), and add:

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

Continue this pattern for the remaining coefficients:

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

The last number (-24) is the remainder. Since it's not zero, x = 1 is not a root of the equation. Bummer!

Testing x = -1

Let's try x = -1. Set up the synthetic division again:

-1 | 1  6  -3  -24  -4
   |        
   ----------------------

Perform the synthetic division steps:

-1 | 1  6  -3  -24  -4
   | -1 -5   8  16
   ----------------------
   1  5 -8 -16 -20

The remainder is -20, so x = -1 is also not a root. Don't lose hope, guys! We've got more possible roots to test.

Testing x = 2

Now, let's try x = 2:

2 | 1  6  -3  -24  -4
  |        
  ----------------------

Perform the synthetic division:

2 | 1  6  -3  -24  -4
  |  2 16  26   4
  ----------------------
  1  8 13   2   0

Eureka! The remainder is 0. This means x = 2 is a root of our equation! We've found one! 🎉

So, from part (b), we've identified that x = 2 is an actual root of the equation.

c) Solving the Equation Using the Root from Part (b)

Okay, we've found one root (x = 2). That's a great start! Now, we need to use this information to solve the rest of the equation. Remember that synthetic division we just did? The numbers in the bottom row (excluding the remainder) are the coefficients of the quotient polynomial. In this case, the quotient is:

$x^3 + 8x^2 + 13x + 2$

Since we divided the original quartic polynomial (degree 4) by a linear factor (x - 2), the quotient is a cubic polynomial (degree 3). Our original equation can now be rewritten as:

$(x - 2)(x^3 + 8x^2 + 13x + 2) = 0$

We already know that x = 2 is a solution from the (x - 2) factor. Now, we need to solve the cubic equation:

$x^3 + 8x^2 + 13x + 2 = 0$

Solving cubic equations can be a bit tricky, but we can use the Rational Root Theorem again! Let's find the possible rational roots for this cubic equation:

• Factors of the constant term (2): ±1, ±2
• Factors of the leading coefficient (1): ±1

Possible Rational Roots: ±1, ±2

We can use synthetic division again to test these roots. Let's try x = -2:

-2 | 1   8   13   2
   | -2 -12  -2
   ----------------
   1   6    1   0

The remainder is 0, so x = -2 is a root of the cubic equation! The quotient is now a quadratic equation:

$x^2 + 6x + 1 = 0$

Our equation is now factored as:

$(x - 2)(x + 2)(x^2 + 6x + 1) = 0$

We have two roots: x = 2 and x = -2. Now, we need to solve the quadratic equation:

$x^2 + 6x + 1 = 0$

This quadratic doesn't factor easily, so we'll use the quadratic formula:

x = rac{-b ± \sqrt{b^2 - 4ac}}{2a}

Where a = 1, b = 6, and c = 1.

Plugging in the values:

x = rac{-6 ± \sqrt{6^2 - 4(1)(1)}}{2(1)} x = rac{-6 ± \sqrt{36 - 4}}{2} x = rac{-6 ± \sqrt{32}}{2} x = rac{-6 ± 4\sqrt{2}}{2} $x = -3 ± 2\sqrt{2}$

So, the two roots from the quadratic equation are:

• $x = -3 + 2\sqrt{2}$
• $x = -3 - 2\sqrt{2}$

Finally, we have all four roots of the original quartic equation:

Solutions: $x = 2$, $x = -2$, $x = -3 + 2\sqrt{2}$, $x = -3 - 2\sqrt{2}$

Conclusion

There you have it, guys! We've successfully solved the quartic equation $x^4 + 6x^3 - 3x^2 - 24x - 4 = 0$ by using the Rational Root Theorem, synthetic division, and the quadratic formula. It might seem like a lot of steps, but breaking it down makes it manageable. Remember, the key is to take it one step at a time, and don't be afraid to try different approaches. Keep practicing, and you'll become a polynomial-solving pro in no time! 😉