Problem 43 - Precalculus Mathematics for Calculus

Question

Find a polynomial with integer coefficients that satisfies the given conditions. \(T\) has degree \(4,\) zeros \(i\) and \(1+i,\) and constant term 12.

Step-by-Step Solution

Answer

The polynomial is \(6x^4 - 12x^3 + 18x^2 - 12x + 12\).

Step 1: Identify all zeros

The given zeros are \(i\) and \(1+i\). Since polynomial coefficients must be real, any complex zeros must occur in conjugate pairs. Hence, the zeros involving \(i\) are \(i\) and \(-i\), and the zeros involving \(1+i\) are \(1+i\) and \(1-i\). Thus, the zeros of the polynomial are \(i, -i, 1+i, 1-i\).

Step 2: Write polynomial in factored form using zeros

The polynomial can be written in its factored form using the zeros: \((x - i)(x + i)(x - (1+i))(x - (1-i))\).

Step 3: Simplify each pair of complex factors

First, simplify \((x - i)(x + i)\): \((x - i)(x + i) = x^2 + 1\), since \((a-b)(a+b) = a^2 - b^2\) and \(i^2 = -1\).Next, simplify \((x - (1+i))(x - (1-i))\): \((x - 1 - i)(x - 1 + i) = ((x - 1)^2 - i^2)\). This becomes \((x - 1)^2 + 1\) because \(i^2 = -1\).

Step 4: Expand the final form of the polynomial

Use the polynomials from the previous step: \( (x^2 + 1)((x - 1)^2 + 1) = (x^2 + 1)(x^2 - 2x + 2) \).Expand: first, expand \( (x - 1)^2 = x^2 - 2x + 1 \), so \( (x^2 + 1)(x^2 - 2x + 1 + 1) = (x^2 + 1)(x^2 - 2x + 2) \).

Step 5: Distribute to express the polynomial

Using distribution, \[ \begin{align*} P(x) & = (x^2 + 1)(x^2 - 2x + 2) \ & = x^2(x^2 - 2x + 2) + 1(x^2 - 2x + 2) \ & = x^4 - 2x^3 + 2x^2 + x^2 - 2x + 2 \ & = x^4 - 2x^3 + 3x^2 - 2x + 2. \end{align*} \]

Step 6: Adjust the constant term to 12

The current polynomial is \(x^4 - 2x^3 + 3x^2 - 2x + 2\). To make the constant term \(12\), multiply the entire polynomial by \(6\) since \(2 \times 6 = 12\) is the desired constant. Thus, the polynomial becomes: \[ P(x) = 6x^4 - 12x^3 + 18x^2 - 12x + 12 \].

Key Concepts

Complex Conjugate Pairs

When dealing with polynomials that have real coefficients and some complex zeros, we need to look at complex conjugate pairs. Complex numbers are in the form of having both real and imaginary parts, like \(a + bi\). Here, \(i\) is the imaginary unit.
In our exercise, the zeros include the numbers \(i\) and \(1+i\). Since the coefficients of the polynomial must be real, each complex zero should come with its conjugate. The complex conjugate of \(a+bi\) is \(a-bi\). Therefore, for every complex zero in our polynomial, there must exist a conjugate zero. For instance:

• \(i\) has \(-i\) as its conjugate pair.
• \(1+i\) has \(1-i\) as its conjugate pair.

This concept ensures that the product of each pair results in a real number.

Real Coefficients

Polynomials are mathematical expressions consisting of variables and coefficients. Coefficients are the numbers in front of the variables. For the polynomial to have real coefficients, it means every number that is not multiplying \(i\) is a real number. That’s why we consider complex conjugate pairs.
Complex numbers alone don’t result in real coefficients. By pairing each with their complex conjugate, we eliminate the imaginary parts through simplification. This is why having real coefficients is crucial; it maintains the polynomial’s characteristics over the real number field, making it understandable and usable in real-world applications such as physics and engineering.

Factored Form

In algebra, factored form helps us express a polynomial as a product of its factors. This can be particularly useful for finding polynomial zeros and simplifying expressions.
In our exercise, once the zeros are identified as \(i, -i, 1+i,\) and \(1-i\), the polynomial is stated in factored form as:

• \((x - i)(x + i)(x - (1+i))(x - (1-i))\)

This structure breaks down the polynomial to its basic roots, providing a clear view of how each zero contributes to the polynomial's form. Furthermore, it makes it easier to simplify and manipulate later on.

Zeros of Polynomial

The zeros of a polynomial are essentially the values of \(x\) for which the polynomial evaluates to zero. Finding the zeros of a polynomial is like solving an equation and represents where its graph crosses the x-axis.
In this exercise, we found the zeros provided and those required by the complex conjugate rule. Here’s how these zeros appear:

• \(i\)
• \(-i\)
• \(1+i\)
• \(1-i\)

These zeros were then used to build the factored form. When multiplied and simplified, these reflect the original polynomial form. Knowing zeros help in graphing and understanding the usability of the polynomial in problem-solving.