Bestimmen Sie s?mtliche reellen und komplexen Lösungen der folgenden Gleichungen: a) \(x^{3}-x^{2}+4 x-4=0\) (Horner-Schema verwenden) b) \(x^{4}-2 x^{2}-3=0\)

Horner's method is a systematic way to simplify polynomial equations and find their roots. It's especially useful for polynomials with higher degrees. To apply Horner's method:

First, you guess a potential root, usually starting with easy numbers like 0, 1, or -1. Then, you use this root to simplify the polynomial recursively. If the guess gives a remainder of zero, congratulations! You've found a root. Now, you can factor the polynomial by dividing it by \(x - root\). This simplifies the polynomial, allowing further factorization.

For example, given the polynomial \(x^3 - x^2 + 4x - 4 = 0\), we guess that \(x = 1\) might be a root. We substitute 1 into the polynomial and find that it indeed equals zero:

\(1^3 - 1^2 + 4(1) - 4 = 1 - 1 + 4 - 4 = 0\)

This confirms \(x = 1\) as a root. Next, by using synthetic division, we divide the polynomial by \(x - 1\). The resulting quotient is a simpler polynomial of lower degree, making it easier to solve further.

Synthetic division is a simple and efficient way to divide polynomials, especially when one of the roots is known. Unlike regular polynomial division, synthetic division only works with coefficients, making the process quick.

To carry out synthetic division, write down the coefficients of the polynomial and the known root. Then, repeatedly multiply and add to simplify the polynomial. Continuing with our example from above:

The polynomial \(x^3 - x^2 + 4x - 4\) and the root \(x = 1\) gives coefficients [1, -1, 4, -4]. We set up the synthetic division as follows:

• Write down the coefficients: 1, -1, 4, -4


• Write the root value (1) on the left.


• Begin the process by dropping the first coefficient (1) straight down.

Now, multiply the root (1) by this dropped value (1) and add it to the next coefficient (-1), continuing this process until all coefficients are considered. This results in a new simplified polynomial form:

The quotient from synthetic division is \(x^2 + 4\). Combined with the known root factor, the original polynomial is now expressed as \( (x-1)(x^2 + 4) = 0 \).

Quadratic equations are polynomials of degree 2 and take the form \(ax^2 + bx + c = 0\). Finding the solutions involves factoring, completing the square, or using the quadratic formula:

\x = \frac{−b± \sqrt{b^2 - 4ac}}{2a}.

In the Exercise given above, we simplify the quadratic equation \(x^2 + 4 = 0\) as part of solving \(x^3 - x^2 + 4x - 4 = 0\). Factoring is not an option here, so we solve it by taking the square root:

\[\begin{equation} x^2 = -4 \Rightarrow x = ±2i \ \sqrt {4} \end{equation}\]

Certain quadratic equations may yield real roots, while others yield complex roots when the discriminant \(b^2 - 4ac\) is negative.

Understanding the nature of solutions is crucial in solving polynomial equations. Solutions can be either real or complex, and they tell us about the polynomial's behavior.

Real solutions are straightforward and lie on the number line. For instance, in the equation \(x^4 - 2x^2 - 3 = 0\), the real solutions found were \pm \sqrt{3}\.

Complex solutions involve imaginary numbers and occur when taking the square root of a negative number. These solutions are represented with the imaginary unit \(i\), where \(i = \sqrt{-1}\). From the equation \(x^3 - x^2 + 4x - 4 = 0\), we found the complex solutions \(2i\ and \-2i\).

In summary:

• Real solutions can be graphed directly on the number line.


• Complex solutions include an imaginary component and exist as pairs when derived from polynomials with real coefficients.

Understanding when and how these solutions appear helps in fully solving polynomial equations and understanding their complete set of roots.