Solving Quartic Equations: A Case Study with the Polynomial x4 - 4x3 - 8x2 - 8x 4 0
In this article, we will explore the process of solving a quartic equation, specifically the polynomial equation x4 - 4x3 - 8x2 - 8x 4 0. We will discuss the methods used to factor and solve such equations and provide a step-by-step guide to finding the roots of the given polynomial.
Introduction to Quartic Equations
A quartic equation is a polynomial equation of the fourth degree. In general form, it is written as:
a4x4 a3x3 a2x2 a1x a0 0
where a4 is different from zero. These equations can be quite complex and may not always be easily factorable. In this case, we are dealing with the specific polynomial:
x4 - 4x3 - 8x2 - 8x 4 0
Initial Observations and Root Testing
The first step in solving a polynomial equation is to look for any obvious roots. In this case, let's test some potential rational roots using the Rational Root Theorem. The Rational Root Theorem states that any rational root of the polynomial, expressed as a fraction p/q, must have p as a factor of the constant term a0 and q as a factor of the leading coefficient a4.
The possible rational roots for our polynomial are the factors of 4 (1, -1, 2, -2, 4, -4). Let's test these values:
x 2:
24 - 4(2)3 - 8(2)2 - 8(2) 4 16 - 32 - 32 - 16 4 -48 ≠ 0
x 1:
14 - 4(1)3 - 8(1)2 - 8(1) 4 1 - 4 - 8 - 8 4 -15 ≠ 0
x 0:
04 - 4(0)3 - 8(0)2 - 8(0) 4 4 ≠ 0
x 4:
44 - 4(4)3 - 8(4)2 - 8(4) 4 256 - 256 - 128 - 32 4 -100 ≠ 0
Since none of these simple rational values are roots, we need to consider other methods such as factoring or numerical methods.
Factoring the Polynomial
Notice that the given polynomial can be rewritten in a form that suggests a simpler factorization. Let's rewrite it as:
x4 - 4x3 - 8x2 - 8x 4 x2(x2 - 2x - 2) 4(1 - 2)
We can then factor the polynomial as:
x2 - 2x - 2 0
Solving the Quadratic
Now, we solve the quadratic equation x2 - 2x - 2 0 using the quadratic formula:
x u00bd(-b ± u221A(b2 - 4ac))
where a 1, b -2, and c -2. Substituting these values, we get:
x u00bd(2 ± u221A((-2)2 - 4(1)(-2)))
x u00bd(2 ± u221A(4 8))
x u00bd(2 ± u221A12)
x u00bd(2 ± 2u221A3)
Therefore, the roots are:
x 1 ± u221A3
Conclusion
The roots of the original polynomial x4 - 4x3 - 8x2 - 8x 4 0 are:
x 1 u221A3 and x 1 - u221A3, each with a multiplicity of 2.
In summary, solving a quartic equation often requires a combination of methods, including recognizing patterns and factoring the polynomial. This case study demonstrates the process of solving a complex quartic equation through polynomial factorization and the quadratic formula.