Solving X^4+6x^3-33x^2-46x+72=0 A Step-by-Step Guide

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Let's dive into solving a fascinating quartic equation, guys! We've got this equation staring back at us:

x^4 + 6x^3 - 33x^2 - 46x + 72 = 0

And our mission, should we choose to accept it (and we totally do!), is to find all the possible values of x that make this equation true. It looks intimidating, I know, but don't worry! We'll break it down step by step. Quartic equations, those polynomial equations with the highest power of x being 4, can seem like monsters at first glance. Unlike our friendly quadratic equations that we can tame with the quadratic formula, quartics require a more strategic approach. But fear not! We’re going to arm ourselves with the Rational Root Theorem, some clever factorization techniques, and a dash of algebraic manipulation to conquer this mathematical beast.

The Rational Root Theorem: Our First Clue

So, where do we even begin? Well, the Rational Root Theorem is like our trusty compass in the wilderness of polynomial equations. This theorem gives us a list of potential rational roots – that is, roots that can be expressed as fractions. How cool is that? It doesn’t guarantee that these will be the actual roots, but it narrows down our search considerably.

The theorem states that if a polynomial equation with integer coefficients has rational roots, they must be of the form p/q, where p is a factor of the constant term (the term without any xs) and q is a factor of the leading coefficient (the coefficient of the x^4 term).

In our case:

  • The constant term is 72. Its factors (p) are ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±24, ±36, and ±72.
  • The leading coefficient is 1 (the coefficient of x^4). Its factors (q) are ±1.

This makes our list of potential rational roots (p/q) look like this: ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±24, ±36, and ±72. Okay, it's still a hefty list, but way better than trying out every number in the universe!

Testing the Potential Roots: Trial and Error (But Smart Trial and Error!)

Now comes the part where we put on our detective hats and test these potential roots. We can do this by directly substituting each number into the equation and seeing if it equals zero. If it does, bingo! We've found a root. Alternatively, we can use synthetic division, which is a more efficient way to test roots and also gives us the quotient polynomial (more on that later).

Let's start with the easiest ones: 1 and -1.

  • Testing x = 1: 1^4 + 6(1)^3 - 33(1)^2 - 46(1) + 72 = 1 + 6 - 33 - 46 + 72 = 0. Jackpot! x = 1 is a root.

  • Testing x = -1: (-1)^4 + 6(-1)^3 - 33(-1)^2 - 46(-1) + 72 = 1 - 6 - 33 + 46 + 72 = 80. Nope, -1 is not a root.

We found our first root, x = 1! That's a great start. Remember, since it's a quartic equation, we are expecting to find up to four roots (some might be repeated, and some might be complex numbers, but let's not get ahead of ourselves).

Synthetic Division: Unlocking the Quotient Polynomial

Now that we've found a root, x = 1, we can use synthetic division to simplify our equation. Synthetic division is a nifty shortcut for dividing a polynomial by a linear factor (like x - 1). It helps us reduce the degree of the polynomial, making it easier to solve.

Here’s how synthetic division works in this case:

  1. Write down the coefficients of the polynomial: 1 6 -33 -46 72
  2. Write the root we found (1) to the left.
  3. Bring down the first coefficient (1).
  4. Multiply the root (1) by the number you just brought down (1), and write the result (1) under the next coefficient (6).
  5. Add the numbers in that column (6 + 1 = 7).
  6. Repeat steps 4 and 5 for the remaining coefficients.
1 | 1   6  -33  -46  72
  |     1    7  -26 -72
  ----------------------
    1   7  -26  -72   0

The last number in the bottom row (0) is the remainder. Since it's zero, it confirms that x = 1 is indeed a root. The other numbers in the bottom row (1, 7, -26, -72) are the coefficients of the quotient polynomial. This means that after dividing our original polynomial by (x - 1), we get a new polynomial:

x^3 + 7x^2 - 26x - 72 = 0

Look at that! We’ve gone from a quartic to a cubic equation. Much more manageable, right?

Solving the Cubic Equation: Back to the Rational Root Theorem (Again!) or Factoring

Now we have a cubic equation: x^3 + 7x^2 - 26x - 72 = 0. We can use the Rational Root Theorem again, this time on the cubic polynomial. The factors of -72 are the same as before, and the leading coefficient is still 1, so our list of potential rational roots remains the same.

Let's try a few. We already know 1 isn't a root of the cubic (it's a root of the quartic, but after dividing by (x-1), it's no longer a root of the cubic). Let’s try x = -2:

(-2)^3 + 7(-2)^2 - 26(-2) - 72 = -8 + 28 + 52 - 72 = 0

Success! x = -2 is a root of the cubic equation.

We can use synthetic division again to reduce the cubic to a quadratic:

-2 | 1   7  -26  -72
   |    -2  -10   72
   ------------------
     1   5  -36    0

This gives us the quadratic equation:

x^2 + 5x - 36 = 0

Solving the Quadratic Equation: The Final Stretch

Ah, a quadratic! These are our old friends. We can solve this using the quadratic formula, factoring, or completing the square. Factoring seems easiest here. We need two numbers that multiply to -36 and add up to 5. Those numbers are 9 and -4.

So, we can factor the quadratic as:

(x + 9)(x - 4) = 0

This gives us two more roots: x = -9 and x = 4.

Putting It All Together: The Solution

We've done it! We've found all the roots of the quartic equation:

  • x = 1
  • x = -2
  • x = -9
  • x = 4

Therefore, the solution is x = [1], [-2], [-9], [4].

What a journey! We tackled a quartic equation, used the Rational Root Theorem, synthetic division, and factoring to find all its roots. Give yourselves a pat on the back, guys! You've conquered a challenging math problem.

Let's dig a little deeper into one of the key tools we used: the Rational Root Theorem. This theorem is like a treasure map for finding potential solutions to polynomial equations, particularly those with integer coefficients. It doesn't guarantee we'll find a root, but it significantly narrows down the possibilities, saving us a ton of time and effort.

The Essence of the Theorem

At its heart, the Rational Root Theorem provides a systematic way to identify potential rational roots. A rational root, remember, is simply a root that can be expressed as a fraction p/q, where p and q are integers. The theorem states:

If a polynomial equation with integer coefficients has rational roots, then these roots must be of the form p/q, where:

  • p is a factor of the constant term of the polynomial (the term without any xs).
  • q is a factor of the leading coefficient of the polynomial (the coefficient of the term with the highest power of x).

Let’s break this down with an example.

Consider a general polynomial equation:

a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 = 0

Where a_n, a_{n-1}, ..., a_1, and a_0 are integer coefficients. According to the Rational Root Theorem:

  • The possible values for p are the factors (both positive and negative) of a_0 (the constant term).
  • The possible values for q are the factors (both positive and negative) of a_n (the leading coefficient).
  • Any rational root of the polynomial must be in the form of p/q, where p and q are chosen from these sets of factors.

Why Does It Work? The Underlying Logic

The magic of the Rational Root Theorem might seem a bit mysterious at first, but it stems from some fundamental principles of polynomial algebra. Here's a simplified way to think about the reasoning behind it.

Let's say we have a polynomial equation with integer coefficients and a rational root p/q (in its simplest form, meaning p and q have no common factors other than 1). This means that when we substitute p/q for x in the polynomial equation, the result should be zero.

When we substitute and manipulate the equation algebraically, we can show that:

  1. p must be a factor of the constant term (a_0).
  2. q must be a factor of the leading coefficient (a_n).

The full proof involves a bit more algebraic manipulation, but this gives you the general idea. The theorem essentially arises from the relationships between the coefficients of the polynomial and its rational roots.

Practical Application: Finding Potential Roots

The real power of the Rational Root Theorem lies in its practical application. It gives us a concrete list of potential rational roots to test. Instead of randomly guessing numbers, we have a focused set of possibilities derived directly from the polynomial's coefficients.

Let's revisit our example equation from earlier:

x^4 + 6x^3 - 33x^2 - 46x + 72 = 0

Here:

  • The constant term is 72, so the possible values for p are the factors of 72: ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±24, ±36, ±72.
  • The leading coefficient is 1, so the possible values for q are the factors of 1: ±1.

Therefore, the potential rational roots (p/q) are: ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18, ±24, ±36, ±72. It's a list, but a finite and manageable one! We can now systematically test these numbers (using substitution or synthetic division) to see if they are actual roots.

Important Considerations and Limitations

It's important to keep in mind the limitations of the Rational Root Theorem:

  • It only identifies potential rational roots. A polynomial equation might have real roots that are irrational (like √2) or complex roots, which the Rational Root Theorem won't help us find directly.
  • The theorem doesn't guarantee that any of the potential rational roots are actual roots. We still need to test them.
  • If the leading coefficient is not 1, the list of potential rational roots can become quite long, as we need to consider all the factors of both the constant term and the leading coefficient. However, it’s still a much more efficient approach than random guessing.

Despite these limitations, the Rational Root Theorem remains a valuable tool in our polynomial-solving arsenal. It gives us a starting point, a structured way to approach finding solutions, and can often lead us to one or more rational roots, which can then be used to further simplify the equation.

Synthetic division is a fantastic shortcut technique in algebra, particularly useful when dealing with polynomials. It's a streamlined method for dividing a polynomial by a linear factor of the form (x - c), where c is a constant. But its usefulness doesn't stop there! Synthetic division is also a powerful tool for finding roots of polynomials and for evaluating polynomial functions at specific values. Let’s break down what makes it so effective and how to use it.

The Mechanics of Synthetic Division: A Step-by-Step Guide

The process of synthetic division might seem a bit like a magic trick at first, but it's built on solid algebraic principles. Here’s how it works:

  1. Set up the division:

    • Write the coefficients of the polynomial you're dividing in a row, making sure to include 0 as a placeholder for any missing terms (e.g., if you're dividing x^4 + 2x^2 - 1 by something, you'd write the coefficients as 1 0 2 0 -1).
    • Write the value of c (from the divisor x - c) to the left.
    • Draw a horizontal line below the coefficients, leaving space for a row of numbers below it.

  2. Bring down the first coefficient:

    • Bring the first coefficient down below the line.

  3. Multiply and add:

    • Multiply the value of c by the number you just brought down.
    • Write the result under the next coefficient.
    • Add the numbers in that column (the coefficient and the result of the multiplication).
    • Write the sum below the line.

  4. Repeat:

    • Repeat the