Problem 25 - College Algebra

Question

In Exercises \(23-28\), factor each polynomial: a. as the product of factors that are irreducible over the rational numbers. b. as the product of factors that are irreducible over the real numbers. c. in completely factored form involving complex nonreal, or imaginary, numbers. $$ x^{4}+x^{2}-6 $$

Step-by-Step Solution

Answer

The factored form over the rational numbers is \((x^2 - 2)(x^2 + 3)\), over the real numbers is \((x^2 - 2)(x^2 + 3)\), and over the complex numbers is \((x - \sqrt{2})(x + \sqrt{2})(x - iz)(x + iz)\)

Step 1: Rewrite the polynomial as a quadratic equation

Observe that the exponent of the first term is twice the exponent of the second term. Hence, this can be rewritten as a quadratic equation. Let \(y = x^2\). The equation then becomes \(y^2 + y - 6\).

Step 2: Factorize the quadratic equation

Factor the quadratic equation using the formula \(y^2 + y - 6 = (y - a)(y - b)\) where \(a\) and \(b\) are the root of the equation. Solve for \(a\) and \(b\) by the method of factorisation to get \(y^2 + y - 6 = (y - 2)(y + 3)\)

Step 3: Substitute \(y = x^2\) back into the equation

This results in \((x^2 - 2)(x^2 + 3)\) which is the factored form over the rational numbers.

Step 4: Factorize over the real numbers

The term \(x^2 - 2\) can be factorized over the real numbers. However, the term \(x^2 + 3\) can't be factorized over the real numbers. Hence, the factored form over the real numbers remains as \((x^2 - 2)(x^2 + 3)\)

Step 5: Factorize over the complex numbers

The term \(x^2 + 3\) can be factorized over the complex numbers such that \(x^2 + 3 = (x - \sqrt{-3})(x + \sqrt{-3}) = x - iz)(x + iz)\),where \(i\) is the imaginary unit such that \(i^2 = -1\). Therefore, the factored form over the complex numbers is \((x - \sqrt{2})(x + \sqrt{2})(x - iz)(x + iz)\)

Key Concepts

Irreducible Factors Over Rational Numbers

When we discuss polynomial factoring in algebra, the term 'irreducible factors over rational numbers' often comes up. This refers to factors of a polynomial that cannot be broken down further using rational numbers, which are numbers that can be expressed as a fraction of two integers.\r\rFor the given exercise, where we have the polynomial \( x^{4}+x^{2}-6 \), the first step is to recognize any hidden quadratic form. By letting \( y = x^2 \), it transforms into a quadratic equation, \( y^2 + y - 6 \). This is a clever technique to simplify the polynomial and make it easier to factorize.\r\rNext, we factorize the quadratic equation into \( (y - 2)(y + 3) \). It's important to mention the significant role the fundamental theorem of algebra plays here, indicating that every non-zero single-variable polynomial has at least one complex root. However, in this context, we're restricted to rational numbers, so we're only interested in rational roots.\r\rReplacing \( y \) with \( x^2 \) again, we get \( (x^2 - 2)(x^2 + 3) \). These are the irreducible factors over rational numbers since neither \( x^2 - 2 \) nor \( x^2 + 3 \) can be factored further using only rational numbers.\r\rThis concept is crucial because it lays the foundation for advanced algebra, where understanding the limitations of rational numbers when factoring polynomials becomes a key skill.

Quadratic Equations

Quadratic equations are fundamental to algebra and appear in numerous forms across various exercises, including the exercise at hand. A quadratic equation is typically expressed in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \) and \( c \) are coefficients and \( a \) cannot be zero.\r\rThe equation from our exercise, after the substitution \( y = x^2 \), becomes a quadratic equation, \( y^2 + y - 6 \), and forms the basis of our factoring process. Factoring involves finding two binomials that when multiplied together give back the original quadratic equation, which in our case are \( (y - 2)(y + 3) \).\r\rFurthermore, quadratic equations can be solved using various methods such as factoring, completing the square, and using the quadratic formula. Understanding these methods are immensely valuable as they provide tools to tackle a wide range of problems in algebra. The ability to manipulate and factorize quadratic equations opens the door to solve for unknowns and apply these skills in more complex scenarios in mathematics.

Complex Numbers in Algebra

Complex numbers expand the field of algebra substantially. They are numbers of the form \( a + bi \), where \( a \), and \( b \) are real numbers, and \( i \) is the imaginary unit with the property that \( i^2 = -1 \). In the context of the provided exercise, after factoring over the rational and real numbers, we get \( (x^2 - 2)(x^2 + 3) \).\r\rOver the real numbers, the term \( x^2 + 3 \) cannot be factored further, as it does not have real roots. However, when we factor over complex numbers, we embrace the fact that every polynomial equation has a root in the complex number system. This is sometimes referred to as the fundamental theorem of algebra.\r\rThus, \( x^2 + 3 \) can be expressed as \( (x - i\sqrt{3})(x + i\sqrt{3}) \) using complex numbers. It’s important to understand that complex numbers are not