Problem 10 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-10: A polynomial \(f(x)\) with real coefficients and leading coefficient 1 has the given zero(s) and degree. Express \(f(x)\) as a product of linear and quadratic polynomials with real coefficients that are irreducible over \(\mathbb{R}\). $$ 0,3 i, 4+i ; \quad \text { degree } 5 $$

Step-by-Step Solution

Answer

The polynomial is \( f(x) = x(x^2 + 9)(x^2 - 8x + 17) \).

Step 1: Identify Real and Complex Conjugates

The given zeros are 0, \(3i\), and \(4+i\). The polynomial has real coefficients, which means complex roots must come in conjugate pairs. The zeros are 0, \(3i\), \(-3i\), \(4+i\), and \(4-i\).

Step 2: Understand the Polynomial Degree

The degree of the polynomial is 5, meaning it should be expressed as a product polynomial with roots corresponding to the 5 zeros identified: 0, \(3i\), \(-3i\), \(4+i\), and \(4-i\).

Step 3: Form the Polynomial with Real Coefficients

Start forming quadratic factors for the complex zeros and a linear factor for the real zero: \[(x)(x^2 + 9)(x^2 - 8x + 17)\].These factors correspond to roots 0, \(3i\) and \(-3i\), and \(4+i\) and \(4-i\).

Step 4: Verify Irreducibility

Check each factor to ensure it is irreducible over \(\mathbb{R}\): - \(x\) is linear.- \(x^2 + 9\) has no real roots (discriminant \(-36\) < 0).- \(x^2 - 8x + 17\) has no real roots (discriminant \(-12\) < 0).All factors are irreducible over \(\mathbb{R}\).

Step 5: Present the Final Polynomial

The polynomial, expressed as a product of irreducible linear and quadratic polynomials with real coefficients, is:\[ f(x) = (x)(x^2 + 9)(x^2 - 8x + 17) \].

Key Concepts

Complex Zeros

When working with polynomial equations, it's important to understand the concept of complex zeros. A zero of a polynomial is a value where the function equals zero. Complex zeros are solutions in the form of a complex number, which means they include both a real and an imaginary part. For example, in the problem, we have zeros such as \(3i\) and \(4+i\). Complex numbers are special because, when dealing with polynomials that have real coefficients, these complex solutions often come in pairs called conjugates. If \(a + bi\) is a root, then its conjugate \(a - bi\) must also be a root to ensure the polynomial remains with real coefficients. This means if you have a zero like \(3i\), you automatically know \(-3i\) is also a zero.

Real Coefficients

In polynomial equations, "real coefficients" refer to the numbers that multiply the variable terms in the equation being real numbers. For instance, when a polynomial is expressed as \(ax^n + bx^{n-1} + \ldots + c\), all coefficients \(a, b, \ldots, c\) are real numbers. Working with real coefficients leads to particular behavior with complex zeros. Since the polynomial in the exercise has real coefficients, every non-real zero must have its complex conjugate also as a zero. This ensures that when the polynomial is multiplied and simplified, the result remains with real numbers and does not include any imaginary components.

Degree of Polynomial

The degree of a polynomial is an important concept that roughly corresponds to the 'highest power' of the variable present in the equation. It indicates the number of roots (including complex and repeated ones) the polynomial will have. For example, the polynomial in the exercise is given to have a degree of 5. This means the polynomial will have exactly 5 roots, which can be a combination of real and complex numbers. We have identified the roots as 0, \(3i\), \(-3i\), \(4+i\), and \(4-i\). The degree helps us arrange all these zeros into the polynomial expression correctly, confirming that we’re covering all the necessary roots.

Irreducible Polynomials

A polynomial is called irreducible if it cannot be factored into simpler polynomials with real coefficients further. For a polynomial with real coefficients, irreducibility means it can't be expressed as a product of two smaller degree polynomials with real numbers. In the exercise, after identifying the zeros, we factor the polynomial into \[(x)(x^2 + 9)(x^2 - 8x + 17)\]. Each piece is checked for irreducibility. The factors \(x^2 + 9\) and \(x^2 - 8x + 17\) are quadratic and do not have real roots (since their discriminants are less than zero), making them irreducible over \(\mathbb{R}\). This concept ensures that the polynomial is broken down into the simplest irreducible components possible, while still retaining real coefficients.