Solved problems: Constructing polynomials from zeros and an additional point

📚 Constructing Polynomials: A Comprehensive Guide

Constructing a polynomial from its zeros and an additional point involves using the zeros to form factors and then using the point to find the leading coefficient. Here's a detailed explanation:

📜 History and Background

The concept of finding polynomials from their roots dates back to the study of algebraic equations. Mathematicians have long been interested in understanding the relationship between the roots of a polynomial and its coefficients. This problem has practical applications in various fields, including engineering, physics, and computer science.

🔑 Key Principles

• 🌱 Zeros and Factors: If $x = a$ is a zero of a polynomial, then $(x - a)$ is a factor of the polynomial. This is a fundamental concept.
• 🧱 Building the Polynomial: Use each zero to create a factor. Multiply these factors together. This gives you a polynomial with the correct zeros, but possibly the wrong leading coefficient.
• 🎯 Using the Additional Point: Substitute the x and y values of the additional point into your polynomial. Solve for the leading coefficient, often denoted as 'a'.
• ✏️ Final Form: Substitute the value you found for 'a' back into your polynomial. This is your final answer.

🧩 Real-World Examples

Let's walk through some examples to illustrate the process:

Example 1:

Construct a polynomial with zeros at $x = 1$, $x = -2$, and $x = 3$, and passing through the point $(2, -4)$

1. 🔍 Form the Factors: The factors are $(x - 1)$, $(x + 2)$, and $(x - 3)$.
2. 🧱 Initial Polynomial: The polynomial is of the form $f(x) = a(x - 1)(x + 2)(x - 3)$.
3. 🎯 Use the Point: Substitute $(2, -4)$ into the polynomial: $-4 = a(2 - 1)(2 + 2)(2 - 3)$. This simplifies to $-4 = a(1)(4)(-1)$, so $-4 = -4a$.
4. 💡 Solve for a: Solving for $a$, we get $a = 1$.
5. ✏️ Final Polynomial: The polynomial is $f(x) = 1(x - 1)(x + 2)(x - 3) = x^3 - 2x^2 - 5x + 6$.

Example 2:

Construct a polynomial with zeros at $x = -1$ (multiplicity 2) and $x = 2$, and passing through the point $(1, 6)$.

1. 🔍 Form the Factors: Since $x = -1$ has multiplicity 2, the factors are $(x + 1)^2$ and $(x - 2)$.
2. 🧱 Initial Polynomial: The polynomial is of the form $f(x) = a(x + 1)^2(x - 2)$.
3. 🎯 Use the Point: Substitute $(1, 6)$ into the polynomial: $6 = a(1 + 1)^2(1 - 2)$. This simplifies to $6 = a(2)^2(-1)$, so $6 = -4a$.
4. 💡 Solve for a: Solving for $a$, we get $a = -\frac{3}{2}$.
5. ✏️ Final Polynomial: The polynomial is $f(x) = -\frac{3}{2}(x + 1)^2(x - 2) = -\frac{3}{2}(x^3 + 2x^2 + x - 2x^2 - 4x - 2) = -\frac{3}{2}(x^3 - 3x - 2) = -\frac{3}{2}x^3 + \frac{9}{2}x + 3$.

✍️ Conclusion

Constructing a polynomial from its zeros and an additional point is a multi-step process. It requires a good understanding of factors, the zero product property, and how to solve for unknowns. Practice is key to mastering this skill.

🧠 Practice Quiz

1. 🧪 Construct a polynomial with zeros at $x = 0$, $x = 4$, and $x = -3$, passing through the point $(1, -6)$.
2. 🧬 Construct a polynomial with zeros at $x = 2$ (multiplicity 2) and passing through the point $(3, 5)$.
3. 🔢 Construct a polynomial with zeros at $x = -1$, $x = 1$, and $x = 5$, passing through the point $(0, 5)$.