Degree 3 polynomial with zeros of 1, -2, and 3

A degree 3 polynomial with zeros of \(1\), \(-2\), and \(3\) can be written as:

\[
f(x) = a(x - 1)(x + 2)(x - 3)
\]

Here, \(a\) is a constant that can be any non-zero real number. For simplicity, let's assume \(a = 1\). Expanding the polynomial:

1. Multiply \((x - 1)(x + 2)\):
\[
(x - 1)(x + 2) = x^2 + 2x - x - 2 = x^2 + x - 2
\]

2. Now multiply this by \((x - 3)\):
\[
(x^2 + x - 2)(x - 3) = x^3 - 3x^2 + x^2 - 3x - 2x + 6
\]

3. Combine like terms:
\[
f(x) = x^3 - 2x^2 - 5x + 6
\]

So the polynomial is:

\[
f(x) = x^3 - 2x^2 - 5x + 6
\]

If you want to scale the polynomial by a constant \(a\), the general form would be:

\[
f(x) = a(x^3 - 2x^2 - 5x + 6)
\]

Would you like to explore any more details about this? 

Here are 5 related questions:

1. How do you find the leading coefficient of a polynomial?
2. Can a polynomial have complex roots as well?
3. What are the implications of different values of \(a\) in the polynomial?
4. How does the degree of the polynomial relate to the number of roots?
5. How do we find the derivative of this cubic polynomial?

**Tip:** The Fundamental Theorem of Algebra states that a degree \(n\) polynomial has exactly \(n\) roots (including complex and repeated roots).

Learn how to construct and expand a cubic polynomial with given zeros 1, -2, and 3. This detailed solution covers key algebraic concepts, including polynomial expansion and the Fundamental Theorem of Algebra. Suitable for high school students in Grades 10-12.

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