Verified Answer :

  • Use the Rational Root Theorem to identify potential roots for the quartic polynomial.
  • Identify the real roots \(x=2\) and \(x=-1\) by testing candidates.
  • Divide the polynomial by the factors \((x-2)\) and \((x+1)\) to obtain a quadratic quotient.
  • Solve the remaining quadratic equation \(x^2+6=0\) to find the complex roots \(x = \pm i\sqrt{6}\).
  • The final roots are \(\boxed{2, -1, i\sqrt{6}, -i\sqrt{6}}\).

Explanation

  1. Analyzing the Polynomial Equation

The objective is to find the roots of the fourth-degree polynomial equation \(x^4 - x^3 + 4x^2 - 6x - 12 = 0\). We can begin by searching for rational roots using the Rational Root Theorem. According to this theorem, any rational root of a polynomial with integer coefficients must be a factor of the constant term (in this case, \(-12\)) divided by a factor of the leading coefficient (in this case, \(1\)). The potential rational roots are therefore \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6,\) and \(\pm 12\).

  1. Testing Rational Roots

We test these potential roots by substituting them into the polynomial \(P(x) = x^4 - x^3 + 4x^2 - 6x - 12\):

  • For \(x = 2\): \(P(2) = (2)^4 - (2)^3 + 4(2)^2 - 6(2) - 12 = 16 - 8 + 16 - 12 - 12 = 0\). Thus, \(x = 2\) is a root.
  • For \(x = -1\): \(P(-1) = (-1)^4 - (-1)^3 + 4(-1)^2 - 6(-1) - 12 = 1 + 1 + 4 + 6 - 12 = 0\). Thus, \(x = -1\) is also a root.

Since \(x = 2\) and \(x = -1\) are roots, both \((x - 2)\) and \((x + 1)\) are factors of the polynomial. Their product is \((x - 2)(x + 1) = x^2 - x - 2\).

  1. Factoring and Solving for All Roots

Now, we perform polynomial division of \(P(x) = x^4 - x^3 + 4x^2 - 6x - 12\) by the quadratic factor \(x^2 - x - 2\) to find the remaining factors:
\[\frac{x^4 - x^3 + 4x^2 - 6x - 12}{x^2 - x - 2} = x^2 + 6.\]
Setting this quotient to zero, we get \(x^2 + 6 = 0\), which gives \(x^2 = -6\). Taking the square root, we find the remaining roots to be \(x = \pm \sqrt{-6} = \pm i\sqrt{6}\).

  1. Conclusion

Combining all results, the roots of the equation \(x^4 - x^3 + 4x^2 - 6x - 12 = 0\) 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 satisfy the equation exactly. The final set of roots is \(\{2, -1, i\sqrt{6}, -i\sqrt{6}\}\).

Examples

In engineering, identifying the roots of a polynomial is essential for analyzing the stability of a system. A system's "poles"—which are the roots of the denominator of its transfer function—tell engineers if the system will settle down or oscillate wildly when disturbed. Solving a quartic equation like the one above is a foundational exercise for predicting the behavior of complex mechanical or electronic systems.

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