Problem 58 - 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. \(f(x)=x^{4}-3 x^{3}-x^{2}-12 x-20\) (Hint: One factor is \(x^{2}+4 .\) )

Step-by-Step Solution

Answer

(a) Over the rationals: \(f(x)=(x^{2}+4)(x^{2}-3x-5)\), (b) Over the reals: \(f(x)=(x^{2}+4)(x-((3+\sqrt{29})/2))(x-((3-\sqrt{29})/2))\), (c) In completely factored form: \(f(x)=(x+2i)(x-2i)(x-((3+\sqrt{29})/2))(x-((3-\sqrt{29})/2))\).

Step 1: Recognize the polynomial

The polynomial to factorize is \(f(x)=x^{4}-3 x^{3}-x^{2}-12 x-20\). The hint provided suggests that \(x^{2}+4\) is one factor of this polynomial.

Step 1: Factorization over the rationals

To factorize the polynomial over the rationals, divide the given polynomial by the factor provided in the hint. Dividing \(f(x)\) by \(x^{2}+4\) results in \(x^{2}-3x-5\). So the polynomial as a product of factors irreducible over rationals is \(f(x)=(x^{2}+4)(x^{2}-3x-5)\).

Step 2: Factorization over the reals

To factorize over the reals, see if the factors can be broken down further. The factor \(x^{2}+4\) is irreducible over the reals as it doesn't have real roots. For \(x^{2}-3x-5\), use the quadratic formula to find the roots and put it in factor form \( (x-p)(x-q)\) where p and q are roots. Using the quadratic formula we get the roots as \[x=\frac{3\pm\sqrt{29}}{2}\]. Hence, the factorization of the polynomial over the reals is \(f(x)=(x^{2}+4)(x-((3+\sqrt{29})/2))(x-((3-\sqrt{29})/2))\).

Step 3: Completely factored form

As \(x^{2}+4\) doesn't have real roots but has complex roots. Hence, to completely factorize \(f(x)\), convert \(x^{2}+4\) into \((x+2i)(x-2i)\) to get the completely factored form. Therefore, in completely factored form, \(f(x)\) is \(f(x)=(x+2i)(x-2i)(x-((3+\sqrt{29})/2))(x-((3-\sqrt{29})/2))\).

Key Concepts

Factorization Over the Rationals

Factorization over the rationals involves breaking down a polynomial into factors that cannot be further simplified with rational number coefficients. Many students face challenges with this process because it requires recognizing patterns and understanding the properties of rational numbers.

When working with the polynomial given in the exercise, we first apply the hint that suggests \(x^2+4\) is a factor. Since this factor doesn't have rational roots (roots that can be expressed as a rational number), we acknowledge it as irreducible over the rationals. To find the other factor, we divide the polynomial by \(x^2+4\), resulting in \(x^2-3x-5\), which is another irreducible factor over the rationals. Consequently, our polynomial factorized over the rationals is \(f(x)=(x^2+4)(x^2-3x-5)\).

It's crucial to understand that a polynomial is considered fully factored over the rationals when it's expressed as a product of polynomials that can't be factored further without using irrational or complex numbers.

Factorization Over the Reals

When we move to factorization over the reals, we look for factors using real number solutions to polynomials, which may include irrational numbers. In our example, we already have one factor, \(x^2+4\), that is irreducible over the reals because it has no real roots. However, the other factor \(x^2-3x-5\) can be further factored by finding its real roots.

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) on \(x^2-3x-5\), we find that the roots are irrational numbers \(\frac{3 \pm \sqrt{29}}{2}\), leading to the factors \(x - \frac{3 + \sqrt{29}}{2}\) and \(x - \frac{3 - \sqrt{29}}{2}\). Therefore, our factorization over the reals is the product of these factors: \(f(x)=(x^2+4)(x- \frac{3 + \sqrt{29}}{2})(x- \frac{3 - \sqrt{29}}{2})\).

The key takeaway is that real number roots, whether rational or irrational, are used to factorize polynomials over the reals.

Irreducible Polynomials

Understanding irreducible polynomials is essential for effectively factoring polynomials. A polynomial is considered irreducible over a field (such as the rationals or the reals) if it cannot be factored into the product of two non-constant polynomials with coefficients in that field.

In our example, the factor \(x^2+4\) is irreducible over both the rationals and the reals, as it doesn't yield roots that are rational or real numbers, respectively. The only roots for \(x^2+4\) are complex: \(x = \pm 2i\). Thus, it remains as a factor in both the rational and real factorizations. For the completely factored form of the given polynomial, we include complex roots, resulting in \(f(x)=(x+2i)(x-2i)(x- \frac{3 + \sqrt{29}}{2})(x- \frac{3 - \sqrt{29}}{2})\).

Identifying irreducible polynomials is vital as they represent the 'basic building blocks' of polynomial factorization, and understanding them is key to mastering factorization at all levels.