Problem 21 - A Graphical Approach to Precalculus with Limits

Question

Find a polynomial function \(P(x)\) having leading coefficient 1, least possible degree, real coefficients. and the given zeros. \(-3,2,\) and \(i\)

Step-by-Step Solution

Answer

The polynomial function is \(P(x) = x^4 + x^3 - 5x^2 + x - 6\).

Step 1: Identify Given Zeros

The problem provides three zeros: 1. \(-3\) 2. \(2\) 3. \(i\). Note that since the polynomial must have real coefficients, the complex zero \(i\) suggests that its conjugate \(-i\) must also be a zero.

Step 2: Set Up the Polynomial with All Zeros

Since the polynomial must include \(i\) and \(-i\) to ensure real coefficients, the zeros of the polynomial are \(-3\), \(2\), \(i\), and \(-i\). Thus, the polynomial can be written in factored form as:\[ P(x) = (x + 3)(x - 2)(x - i)(x + i) \]

Step 3: Simplify Complex Factors

The complex factors \((x-i)(x+i)\) can be simplified using the difference of squares:\[ (x - i)(x + i) = x^2 + 1 \]

Step 4: Expand the Polynomial

Now substitute the simplified complex factor back into the polynomial:\[ P(x) = (x+3)(x-2)(x^2+1) \]Next, expand the polynomial by first expanding \((x+3)(x-2)\):\[ (x+3)(x-2) = x^2 - 2x + 3x - 6 = x^2 + x - 6 \]Now multiply by \((x^2+1)\):\[ (x^2 + x - 6)(x^2 + 1) = x^4 + x^3 - 6x^2 + x^2 + x - 6 \]Combine like terms:\[ P(x) = x^4 + x^3 - 5x^2 + x - 6 \]

Step 5: Verify Leading Coefficient

Ensure the leading coefficient of the polynomial is 1, which it is (\(x^4\) term has a coefficient of 1). So the polynomial meets the requirement of having a leading coefficient of 1.

Key Concepts

Real Coefficients

In polynomial equations, real coefficients are essential if the polynomial is expected to model real-world phenomena. Real coefficients simply mean that all the numbers in front of each term are real numbers. Real numbers can be any number that is not imaginary, including whole numbers, fractions, and decimals.
For a polynomial to have real coefficients, the presence of complex numbers in the zeros requires particular attention. When a polynomial has complex roots, these roots must occur in conjugate pairs. A complex conjugate of a complex number has the same real part but the opposite imaginary part.
For example, given a root of a polynomial, like the number \( i \) which is imaginary, its complex conjugate would be \( -i \). Together, the pair \( i \) and \( -i \) ensure that when multiplied, they result in a real number (in this case, \( x^2 + 1 \)). This ensures that the polynomial maintains real coefficients throughout.

Complex Numbers

Complex numbers are numbers that include both a real part and an imaginary part. The imaginary unit is denoted as \( i \), which is defined as \( i^2 = -1 \). Complex numbers expand the idea of what a number is in mathematics, allowing for the solution of equations that cannot be solved using solely real numbers.
In polynomials, introducing a zero as a complex number necessitates including its conjugate to preserve real coefficients. If you have a polynomial that needs a zero at \( i \), you must also include \( -i \) as zero.
This importance grows in polynomials with real coefficients because when you multiply two conjugate pairs like \( (x - i)(x + i) \), the result is a difference of squares which simplifies to a real-number expression (\( x^2 + 1 \)), neutralizing the imaginary parts.

Difference of Squares

The difference of squares is a mathematical identity that simplifies expressions of the form \((a + b)(a - b) = a^2 - b^2\). This property is particularly useful in algebra when simplifying polynomials that include complex conjugates.
When dealing with polynomial zeros that include complex numbers, such as \( i \) and \( -i \), they can form factors like \((x - i)(x + i)\). This pair is a classic difference of squares since the equation equates to \( x^2 - i^2 \). Because \( i^2 = -1 \), the expression further simplifies to \( x^2 + 1 \), effectively converting a complex expression into a simple polynomial term with real coefficients.
Using the difference of squares removes the complexity added by imaginary units, making the polynomial easier to handle and ensuring all terms remain as real numbers.