What are the roots of the equation?
$0=x^4-x^3+4 x^2-6 x-12$
We start by analyzing the polynomial equation \(x^4 - x^3 + 4x^2 - 6x - 12 = 0\). According to the Rational Root Theorem, any rational root \(p/q\) must have \(p\) as a factor of \(-12\) and \(q\) as a factor of \(1\). The possible candidates are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\). Testing some of these candidates, we find that for \(x = 2\), the expression becomes \(16 - 8 + 16 - 12 - 12 = 0\), so \(x = 2\) is a root. Similarly, for \(x = -1\), the expression becomes \(1 + 1 + 4 + 6 - 12 = 0\), so \(x = -1\) is also a root.
Since \(x = 2\) and \(x = -1\) are roots, \((x - 2)\) and \((x + 1)\) are factors of the polynomial. Their product is \((x - 2)(x + 1) = x^2 - x - 2\). We can now divide the original polynomial \(x^4 - x^3 + 4x^2 - 6x - 12\) by \(x^2 - x - 2\) to find the remaining quadratic factor. Performing this polynomial division:
\[(x^4 - x^3 + 4x^2 - 6x - 12) \div (x^2 - x - 2) = x^2 + 6\]
Setting the quotient to zero, we get \(x^2 + 6 = 0\), which gives \(x^2 = -6\). The solutions to this are \(x = \pm i\sqrt{6}\).
We have identified all four roots of the quartic equation. They are the two real roots \(x = 2\) and \(x = -1\), and the two complex roots \(x = i\sqrt{6}\) and \(x = -i\sqrt{6}\). These roots satisfy the original equation \(x^4 - x^3 + 4x^2 - 6x - 12 = 0\).
Finding roots of higher-degree polynomials is essential in fields like control engineering and physics, where the stability of a system or the behavior of wave functions depends on solving characteristic equations. For example, in designing a stable bridge or an electronic circuit, finding the specific values where the system's characteristic polynomial equals zero helps identify its natural frequencies and predict potential resonance issues. Understanding these roots allows engineers to adjust design parameters to ensure safety and optimal performance.