Problem 28 - 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. \(x^{4}-4 x^{3}+14 x^{2}-36 x+45\) (Hint: One factor is \(\left.x^{2}+9 .\right)\)

Step-by-Step Solution

Answer

a. Over rational numbers, the polynomial remains in the same form: \(x^{4}-4 x^{3}+14 x^{2}-36 x+45\). b. Over real numbers, it can be factored as \((x^{2} - 4x + 5) * (x^{2} + 9)\). c. Over complex numbers, the polynomial can be factored as \((x - 2 -i) * (x - 2 + i)*(x-3i)*(x+3i)\).

Step 1: Factoring The Polynomial Over Rational Numbers

Firstly, look into the given polynomial \(x^{4}-4 x^{3}+14 x^{2}-36 x+45\) and see if there are any common factors that can be taken out. In this case, there are none, and we are given a hint that one of the factors is \(x^{2} + 9\). So, for now, we can leave the polynomial as it is without any further factoring.

Step 2: Factoring The Polynomial Over Real Numbers

Since we don't need to be concerned about factors with rational-coefficients, and also given the hint, we can factorize our polynomial as \((x^{2} - 4x + 5) * (x^{2} + 9)\) for real numbers. Half of this process is to confirm that multiplying these two factors gives us the original polynomial.

Step 3: Factoring The Polynomial Involving Complex Non-Real Or Imaginary Numbers

In the realm of complex numbers, the factor \(x^{2} + 9\) can further be factored to \( (x-3i)(x+3i) \). For the other factor \((x^{2} - 4x + 5)\), the roots are complex with real and imaginary parts. Using the quadratic formula, we find the roots are \(2 \pm i\) which translate into factors \((x - 2 -i)\) and \((x - 2 + i)\). So, the polynomial \(x^{4}-4 x^{3}+14 x^{2}-36 x+45\) can be expressed in completely factored form with complex factors as \((x - 2 -i) * (x - 2 + i)*(x-3i)*(x+3i)\).

Key Concepts

Irreducible Factors Over Rational Numbers

When we talk about factoring polynomials over the rational numbers, we're addressing the process of breaking down a polynomial into factors that cannot be further simplified using only numbers that can be expressed as the quotient of two integers. In mathematical terms, these are called irreducible factors. For example, consider the polynomial from our exercise, \(x^{4}-4 x^{3}+14 x^{2}-36 x+45\). This polynomial doesn't have any common factors and we know that \(x^{2} + 9\) is one of its factors. Over the rational numbers, \(x^{2} + 9\) cannot be factored further because it has no real roots, and thus it is already in its irreducible form.

Here's a tip for identifying irreducible factors over the rational numbers:

• Check if the polynomial can be factored by applying techniques like synthetic division or the factor theorem.
• If no rational root can be found, then polynomials like \(x^{2} + 9\) are already irreducible over the rational numbers.

By not finding a factorization that includes rational numbers, we conclude that the polynomial has been factored as much as it can be with regard to the rational number system.

Factoring Over Real Numbers

Moving from rational numbers to real numbers opens up more possibilities for factoring. Every real polynomial can be factored into linear and/or quadratic irreducible factors. The hint provided tells us that \(x^{2} + 9\) is one of the factors, and it remains irreducible over the real numbers because it doesn't have real roots. However, another part of our polynomial can be factored further.
We discovered that the polynomial can be factored as \(x^{2} - 4x + 5\) times \(x^{2} + 9\). The quadratic \(x^{2} - 4x + 5\) does not have any real number factors because its discriminant, \(4^2 - 4(1)(5)\), is negative.

Understanding Quadratic Factorization:

• The discriminant of a quadratic equation \(ax^{2} + bx + c\) is given by \(b^{2} - 4ac\).
• If the discriminant is positive, there are two distinct real roots, leading to the factorization into two linear factors.
• If the discriminant is zero, there is one real root, corresponding to a repeated linear factor.
• If the discriminant is negative, there are no real roots, and the quadratic is irreducible over the real numbers.

This is why over the real numbers, the factored form remains \(x^{2} - 4x + 5\) times \(x^{2} + 9\).

Complex Number Factors

Lastly, factoring over the complex numbers allows us to break down polynomials into linear factors, regardless of whether they have real roots or not. The world of complex numbers introduces imaginary units, usually denoted as \(i\), where \(i^{2} = -1\). This lets us factor any quadratic polynomial with real coefficients completely.
For our exercise, we notice that the factor \(x^{2}+9\) can indeed be further factored into \(x-3i\) and \(x+3i\), because \(i^{2}\) gives us the -1 required to turn +9 into a perfect square and therefore allows for linear factors to emerge. Similarly, the factor \(x^{2} - 4x + 5\) has complex roots that can be found using the quadratic formula.

Complex Roots:

• When a quadratic equation's discriminant is negative, we'll find complex roots.
• These roots are expressed in the form \(a \pm bi\), where \(a\) is the real part and \(bi\) is the imaginary part.
• Complex roots always come in conjugate pairs, meaning if \(a + bi\) is a root, so is \(a - bi\).

Therefore, the completely factored form of \(x^{4}-4 x^{3}+14 x^{2}-36 x+45\) with complex factors becomes \(x - 2 -i\) times \(x - 2 + i\) times \(x-3i\) times \(x+3i\). This profound concept demonstrates the Fundamental Theorem of Algebra, which states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. This theorem guarantees the complete factorization of polynomials in the realm of complex numbers.