Problem 33 - Precalculus with Calculus Previews

Question

Find a polynomial function \(f\) with real coefficients of the indicated degree that possesses the given zeros. degree \(2 ; 1-6 i\)

Step-by-Step Solution

Answer

The polynomial is \(f(x) = x^2 - 2x + 37\).

Step 1: Identify the Conjugate Pair

Since polynomial coefficients are real, the zeros must include the complex conjugate. Given zero is \(1 - 6i\). The complex conjugate is \(1 + 6i\). Thus, the zeros of the polynomial are \(1 - 6i\) and \(1 + 6i\).

Step 2: Set Up the Factored Form

Using the zeros, the polynomial can be expressed in factored form as \((x - (1 - 6i))(x - (1 + 6i))\).

Step 3: Expand the Factored Form

Expand the expression \((x - (1 - 6i))(x - (1 + 6i))\) using the difference of squares formula \((a-b)(a+b) = a^2 - b^2\), where \(a = x - 1\) and \(b = 6i\).

Step 4: Calculate \((x-1)^2\) and \((6i)^2\)

Compute \((x-1)^2 = x^2 - 2x + 1\) and \((6i)^2 = 36i^2 = -36\).

Step 5: Form the Polynomial

Substitute back into the difference of squares: \((x - 1)^2 - (6i)^2 = x^2 - 2x + 1 + 36 = x^2 - 2x + 37\).

Step 6: Write the Final Polynomial

Thus, the polynomial function \(f(x)\) with the given zeros is \(f(x) = x^2 - 2x + 37\).

Key Concepts

Complex Conjugates

When dealing with polynomials that have real coefficients but complex roots, it's essential to understand the role of complex conjugates. Complex conjugates come in pairs and are found when you change the sign of the imaginary part in a complex number. For example, if you have a complex number like \(1 - 6i\), its conjugate would be \(1 + 6i\). This is important because when a polynomial has real coefficients, any complex zero must have its conjugate also be a zero. This ensures the polynomial remains real-valued.

• Remember, complex conjugates involve flipping the sign of the imaginary part.
• Real-coefficient polynomials guarantee conjugate pairs.

If you have one complex root, the combined effect of including its conjugate ensures that any imaginary components cancel out when you expand the polynomial.

Factorization

Factorization is the process of expressing a polynomial as a product of its factors. These factors are typically linear terms involving the zeros of the polynomial. For the given exercise, once we identify the zeros \(1 - 6i\) and \(1 + 6i\), we set up our polynomial in factored form as \((x - (1 - 6i))(x - (1 + 6i))\). This approach reveals that the roots are also the factors of the polynomial.

• Start by identifying all zeros, including complex conjugates.
• Set each factor as \((x - \, \text{zero})\).

This method ensures we accurately capture all required roots in the polynomial's structure. It simplifies the process of expanding and solving these equations.

Difference of Squares

The difference of squares is a helpful algebraic identity used to simplify expressions where each term is a perfect square and involves a subtraction. The formula \((a-b)(a+b) = a^2 - b^2\) can simplify complex expressions easily.In our specific example, we use complex zeros \((x - (1 - 6i))(x - (1 + 6i))\) to transform the expression into this form. The terms here, \((x-1)^2\) and \(6i^2\), fit perfectly into the formula, allowing us to simplify:

• Compute \((x-1)^2 = x^2 - 2x + 1\)
• Calculate \((6i)^2 = 36i^2 = -36\)

Thus, the equation becomes \((x-1)^2 - (6i)^2 = x^2 - 2x + 1 + 36\), leading to \(x^2 - 2x + 37\) without complex components.

Complex Numbers

Complex numbers are fascinating mathematical constructs that consist of a real part and an imaginary part, typically denoted as \(a + bi\). Here, "\(a\)" represents the real part, and "\(bi\)" is the imaginary component where \(i\) is the imaginary unit satisfying \(i^2 = -1\).Handling polynomials with complex zeros requires understanding of these numbers to ensure all solutions are valid:

• Complex numbers allow for solutions beyond real limits, such as square roots of negative numbers.
• In algebra, account for the imaginary unit with \(i^2\) equating to \(-1\).

In problems similar to our exercise, complex numbers expand the capability of polynomial functions to contain a wider range of solutions.