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The image you've uploaded shows a list of polynomial functions, and the task is to determine the zeros (roots) of these functions. Here's how you find the zeros for each function:

### a) \( f(x) = x^5 - 2x^3 + x \)

This function can be factored as follows:
\[ f(x) = x(x^4 - 2x^2 + 1) = x(x^2 - 1)^2 = x(x-1)^2(x+1)^2 \]
The zeros are \( x = 0, 1, -1 \), with \( x = 1 \) and \( x = -1 \) each being a double root.

### b) \( f(x) = 0.25x^3 + 0.25x^2 - 3x \)

This can be factored by taking out the common term:
\[ f(x) = x(0.25x^2 + 0.25x - 3) \]
Let's factor the quadratic part:
\[ 0.25x^2 + 0.25x - 3 = \frac{1}{4}(x^2 + x - 12) \]
Factoring the quadratic:
\[ \frac{1}{4}(x-3)(x+4) \]
Thus, the zeros are \( x = 0, 3, -4 \).

### c) \( f(x) = 16x^4 - 33x^2 + 16 \)

This is a quadratic in terms of \( y = x^2 \):
\[ 16y^2 - 33y + 16 = 0 \]
Using the quadratic formula for \( y \):
\[ y = \frac{33 \pm \sqrt{1089 - 1024}}{32} = \frac{33 \pm \sqrt{65}}{32} \]
Thus, the zeros for \( x \) are:
\[ x = \pm \sqrt{\frac{33 \pm \sqrt{65}}{32}} \]

### e) \( f(x) = x^4 + 2.6x^2 + 0.69 \)

This is also a quadratic in \( y = x^2 \):
\[ y^2 + 2.6y + 0.69 = 0 \]
Using the quadratic formula for \( y \):
\[ y = \frac{-2.6 \pm \sqrt{6.76 - 2.76}}{2} = \frac{-2.6 \pm \sqrt{4}}{2} = -0.3, -2.3 \]
Since \( y = x^2 \geq 0 \), we discard \( y = -2.3 \) and solve \( y = -0.3 \) (if possible; otherwise, it means there are no real roots).

### g) \( f(x) = x^4 - 3x^2 + 2 \)

Again treating as a quadratic in \( y = x^2 \):
\[ y^2 - 3y + 2 = 0 \]
Factoring:
\[ (y-2)(y-1) = 0 \]
Thus, \( y = 2 \) or \( y = 1 \), leading to \( x = \pm 1, \pm \sqrt{2} \).

Would you like to see the solutions for other functions listed in the image, or would you like details or steps for any specific function among these?

### Related Questions:
1. How can a polynomial be factored when there is no obvious common term?
2. What is the importance of checking the discriminant in quadratic equations?
3. How are multiple roots indicated in a factorization?
4. What methods can be used to solve higher degree polynomials that cannot be easily factored?
5. Can graphing be a useful tool for identifying the roots of a polynomial?

**Tip**: For higher degree polynomials that don't factor easily, numerical methods like Newton's method can be very effective.

Bestimme alle Nullstellen der folgenden Funktionen: f(x) = x^5 - 2x^3 + x, f(x) = 0.25x^3 + 0.25x^2 - 3x, f(x) = 16x^4 - 33x^2 + 16, f(x) = x^4 + 2.6x^2 + 0.69, f(x) = x^4 - 3x^2 + 2

Learn how to find the zeros of polynomial functions like f(x) = x^5 - 2x^3 + x using factorization and the quadratic formula. This detailed explanation covers key algebraic concepts for high school students in Grades 10-12.

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