Problem 25 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

The polynomial function is \(P(x) = x^4 - 7x^3 + 17x^2 - x - 26\).

Step 1: Identify all zeros

The given zeros are \(3 + 2i\), \(-1\), and \(2\). Since the coefficients of the polynomial must be real, the complex zeros come in conjugate pairs. So, the conjugate of \(3 + 2i\), which is \(3 - 2i\), is also a zero of the polynomial.

Step 2: Write the factors from the zeros

Each zero \(a\) of the polynomial corresponds to a factor \((x - a)\). So, the factors for the zeros \(3 + 2i\), \(3 - 2i\), \(-1\), and \(2\) are \((x - (3 + 2i))\), \((x - (3 - 2i))\), \((x + 1)\), and \((x - 2)\).

Step 3: Multiply the complex conjugate factors

The factors \((x - (3 + 2i))\) and \((x - (3 - 2i))\) can be multiplied to produce a polynomial with real coefficients: \[(x - (3 + 2i))(x - (3 - 2i)) = ((x - 3) - 2i)((x - 3) + 2i)\] Using the difference of squares formula, \[(x - 3)^2 - (2i)^2 = (x - 3)^2 + 4 = x^2 - 6x + 13\].

Step 4: Form the polynomial

Combine the multiplied complex conjugate pair with the remaining factors from real zeros:\[P(x) = (x^2 - 6x + 13)(x + 1)(x - 2)\].

Step 5: Expand the polynomial expression

First, multiply \((x + 1)(x - 2)\): \[(x + 1)(x - 2) = x^2 - 2x + x - 2 = x^2 - x - 2\].Then, multiply \((x^2 - 6x + 13)\) by \((x^2 - x - 2)\):\[P(x) = (x^2 - 6x + 13)(x^2 - x - 2)\].Distribute each term:\[x^4 - x^3 - 2x^2 - 6x^3 + 6x^2 + 12x + 13x^2 - 13x - 26\].Combine like terms:\[P(x) = x^4 - 7x^3 + 17x^2 - x - 26\].

Key Concepts

Complex Conjugate

When dealing with complex numbers, there's something called the "complex conjugate." A complex number often consists of a real part and an imaginary part, like the number \(3 + 2i\). The complex conjugate is formed by changing the sign of the imaginary part. So, the conjugate of \(3 + 2i\) is \(3 - 2i\).
This concept is really helpful when you want polynomial functions that have real coefficients.

• Whenever there's a complex zero, its conjugate must also be a zero.
• This ensures the polynomial ends up with real numbers after expansion.

In our exercise, since \(3 + 2i\) is a zero, \(3 - 2i\) must be included to maintain real coefficients. This is a common rule when constructing polynomials with real coefficients.

Real Coefficients

Polynomial functions with real coefficients are quite essential in mathematics. They are expressions where all coefficients are real numbers, meaning there are no imaginary parts. This requirement profoundly influences how we handle complex zeros.

• When a polynomial is required to have real coefficients, any complex zero always comes paired with its complex conjugate.
• This mirroring ensures that when you expand the polynomial, the coefficients for all terms are real numbers.

In the given exercise, the polynomial with zeros including \(3 + 2i\) must also include \(3 - 2i\). This balances the imaginary parts to become real through the difference of squares formula:\(((x - 3) - 2i)((x - 3) + 2i) = (x^2 - 6x + 13)\)This kind of treatment keeps the equations nice and real!

Polynomial Zeros

Zeros of a polynomial are values for which the polynomial evaluates to zero. They are critical because they form the foundation of polynomial equations.

• Each zero gives rise to a factor of the polynomial in the form \((x - a)\), where \(a\) is the zero.
• If you find all the zeros of a polynomial and make a factor for each, you can reconstruct the polynomial completely.

In the exercise, the given zeros were \(3 + 2i, -1,\) and \(2\). A missing element, the conjugate zero \(3 - 2i\), was determined from the requirement for real coefficients. Therefore, each zero corresponds to factors:- \(3 + 2i\) yields \((x - (3 + 2i))\)- \(3 - 2i\) yields \((x - (3 - 2i))\)- \(-1\) yields \((x + 1)\)- \(2\) yields \((x - 2)\)Multiplying these factors gives the polynomial. The zeros guide the entire structure of the polynomial function, and understanding them is key when solving such problems.