Problem 57 - Algebra and Trigonometry Real Mathematics, Real People

Question

Write the polynomial (a) as the product of factors that are irreducible over the rationals, (b) as the product of linear and quadratic factors that are irreducible over the reals, and (c) in completely factored form. \(f(x)=x^{4}-2 x^{3}-3 x^{2}+12 x-18\) (Hint: One factor is \(x^{2}-6 .\) )

Step-by-Step Solution

Answer

(a) \((x^{2}-6)(x^{2}-2x+3)\), (b) \((x^{2}-6)(x-1-i)(x-1+i)\), (c) \((x^{2}-6)(x-1-i)(x-1+i)\)

Step 1: Factoring the Polynomial Over the Rationals

Upon implementing the synthetic division or polynomial division against \(x^{2}-6\), the polynomial is simplified as \((x^{2}-6)(x^{2}-2x+3). This can be considered as the factored form over the rationals.

Step 2: Factoring the Polynomial Over the Reals

\((x^{2}-2x+3)\) is a quadratic expression which can be simplistically factorized using the method of completing the square. Prior to this, it's recognized that \((x^{2}-6)\) cannot be factored anymore, hence, it remains same. Using the formula \(-b\pm\sqrt{b^2-4ac}/2a\), the quadratic factor transforms to \((x-1)^2+(1)^2=(x-1-i)(x-1+i)\). Thus, the polynomial can be represented over the reals as \((x^{2}-6)(x-1-i)(x-1+i)\).

Step 3: Factoring the Polynomial Completely

Since the polynomial is factored entirely over the reals, and all factors are linear, this can be considered as completely factored form.

Key Concepts

Irreducible Polynomials

In mathematics, an irreducible polynomial is a polynomial that cannot be factored into polynomials of lower degrees with coefficients in a given field. This means, over the considered field (like rationals or reals), it doesn't break into simpler polynomial factors. For the polynomial provided in the exercise, the factor \(x^2 - 6\) is an irreducible polynomial over the rational numbers. This is because it cannot be factored further into polynomials with rational coefficients. When working with irreducible polynomials, it's essential to identify what field you're working in, as it influences whether a polynomial is considered irreducible. In the context of rational numbers, any polynomial like \(x^2 - 6\) that doesn't have a simplified rational root remains in its form, maintaining its "irreducible" status.

Factoring over Rationals

Factoring a polynomial over the rationals involves breaking it down into polynomials whose coefficients are rational numbers. This is typically the starting point for factorization. In our exercise, you begin by identifying a factor like \(x^2 - 6\). Using polynomial division methods like synthetic division, the polynomial \(f(x) = x^4 - 2x^3 - 3x^2 + 12x - 18\) is divided by \(x^2 - 6\) leading to another quadratic polynomial \(x^2 - 2x + 3\). This process is vital for simplification and further breakdown of complex polynomials. Remember, when factoring over rationals, each factor must have rational coefficients. Thus, the expression becomes \((x^2 - 6)(x^2 - 2x + 3)\) over the rationals.

Completing the Square

Completing the square is a method used to solve and simplify quadratic equations. It transforms a quadratic expression into a perfect square trinomial by adding and subtracting the necessary term. For the quadratic \(x^2 - 2x + 3\) used in our exercise, completing the square helps us express it in a form that can be easily analyzed over the real numbers. By rewriting the quadratic as \((x - 1)^2 + 1^2\), it shows part of its structure as a sum of squares. This process not only aids in factorization but also paves the way for easier identification of roots or helps in dealing with expressions algebraically. It is an essential skill for students to master, especially in transforming expressions to be used with complex numbers.

Complex Numbers

Complex numbers extend our number system beyond the real numbers, incorporating the imaginary unit \(i\), where \(i^2 = -1\). This extension allows us to factor polynomials that aren't factorable over the reals. In our polynomial exercise, once the expression \(x^2 - 2x + 3\) is completed into a square, it rotates into \((x - 1 - i)(x - 1 + i)\), factoring over the complex numbers. Recognizing complex conjugates like \((a - bi)\) and \((a + bi)\) is crucial, as they assure real coefficients when worked through the complex plane. Using complex numbers in polynomial factorization can elegantly solve and represent polynomials' roots that resist simplification in the real world. As such, they form a significant part of advanced algebra and calculus, providing insights into polynomials' geometrical and numerical properties.