Problem 73 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find a polynomial function that has the given zeros. (There are many correct answers.) \(2,2+\sqrt{5}, 2-\sqrt{5}\)

Step-by-Step Solution

Answer

The polynomial function that has the given zeros is \(x^3 - 6x^2 + 14x - 14\).

Step 1: Convert roots into factors

Since \(2, 2 + \sqrt{5}, 2 - \sqrt{5}\) are roots, it means that \(x - 2, x - (2 + \sqrt{5}), x - (2 - \sqrt{5})\) are factors of the polynomial.

Step 2: Simplify the factors

Simplify the second and third factor to \(x - 2,\; x - 2 - \sqrt{5},\; x - 2 + \sqrt{5}\).\nThis can also be represented as \(x - 2, x - 2 - \sqrt{5} , x - 2 + \sqrt{5}\).

Step 3: Multiply the factors together

Now, multiply these factors to get the polynomial. When you do the calculations, the polynomial will be \( (x - 2) \cdot (x - 2 - \sqrt{5}) \cdot (x - 2 + \sqrt{5}) \), which simplifies to \(x^3 - 6x^2 + 14x - 14\).

Key Concepts

Finding Roots of Polynomials

To understand how to find the roots of polynomials, we first need to grasp what a polynomial is. A polynomial is an algebraic expression that involves a sum of powers in a variable, where each power has a coefficient. The roots of a polynomial are the values of the variable that make the polynomial equal to zero. For a polynomial equation like \(f(x) = 0\), the roots are the solutions, or the values of \(x\), that satisfy this equation.
Identifying the roots is vital because they represent the x-intercepts on the graph of the polynomial. If you have a polynomial and you know some of its roots, you can factor the polynomial into linear factors. This is a reverse approach from how we handle factoring in algebra.

• Each root \(r\) of a polynomial \(f(x)\) translates into a factor \((x - r)\).
• For example, if 2 is a root of the polynomial, then \(x-2\) is a factor.
• Find all factors by converting every root to its linear expression.

In our specific exercise, we have roots \(2, 2 + \sqrt{5}, 2 - \sqrt{5}\). Thus, corresponding factors are \((x - 2), (x - (2 + \sqrt{5})), (x - (2 - \sqrt{5}))\). These factors are crucial for multiplying them back into a full polynomial as shown in the following sections.

Multiplying Polynomials

Multiplying polynomials can be simplified by remembering the distributive property, which states that each term in the first polynomial must be multiplied by each term in the second polynomial. This results in combining like terms in the process to simplify the expression.
In the given exercise, we have three factors derived from the roots: \( (x - 2), (x - (2 + \sqrt{5})), \) and \( (x - (2 - \sqrt{5})) \). Notice how the last two factors can be rewritten using the difference of squares identity \((a-b)(a+b) = a^2 - b^2\).

• First, multiply \((x - (2 + \sqrt{5}))\) and \((x - (2 - \sqrt{5}))\) using the special product formula to get \( [(x-2)^2 - (\sqrt{5})^2] \).
• This simplifies to \( (x-2)^2 - 5 \).
• Next, expand: \((x-2)(x^2 - 4x - 1)\).

By using these strategies, you can develop the product into a simplified polynomial such as \( x^3 - 6x^2 + 14x - 14 \), which represents the polynomial function with the given roots.

Simplifying Algebraic Expressions

Simplifying algebraic expressions is an essential skill in algebra. It involves reducing expressions to their simplest form, making them easier to read and work with. This often requires combining like terms and using algebraic identities.
In the context of the problem provided, after multiplying the factors of the polynomial, you might have intermediate terms that require simplification. Similar terms are combined, and equations are rewritten in a more manageable way.

• Combine like terms: Terms with the same variable and exponent are added or subtracted. For example, \(x^2\) terms are combined with other \(x^2\) terms.
• Use algebraic identities: The identity \((a+b)^2 = a^2 + 2ab + b^2\) is one of the vital tools used in simplification.
• Simplification leads to polynomials like \(x^3 - 6x^2 + 14x - 14\), where all terms are combined to achieve a clear and concise expression.

With practice, identifying and applying these rules becomes second nature, facilitating a smoother algebra-solving experience.