Problem 113 - Introductory Algebra for College Students

Question

Explain why \(x^{2}-1\) is factorable, but \(x^{2}+1\) is not.

Step-by-Step Solution

Answer

The expression \(x^{2}-1\) is factorable because it can be expressed as the product of two binomials with real numbers \((x-1)(x+1)\). On the other hand, \(x^{2}+1\) is not factorable in the set of real numbers because it cannot be expressed as the product of binomials with real coefficients. The roots of \(x^{2}+1\) are imaginary numbers.

Step 1: Understanding Factoring

Factoring is the process of breaking down an expression into a product of simpler expressions. For a quadratic expression \(ax^{2} + bx + c\), it is factorable if there exist real numbers \(p\) and \(q\) such that the expression can be rewritten as \(a(x - p)(x - q)\). The real numbers \(p\) and \(q\) are called the roots of the quadratic expression.

Step 2: Factorizing \(x^{2}-1\)

We observe that \(x^{2}-1\) can be expressed in the form \((x-1)(x+1)\), which means it is factorable and its roots are x = 1 and x = -1.

Step 3: Checking Factorization of \(x^{2}+1\)

The expression \(x^{2}+1\) cannot be written as a product of two binomials with real coefficients. Technically, it has roots, but they are complex: \(x = i\) and \(x = -i\), where \(i\) is the imaginary unit. Therefore, \(x^{2}+1\) is not factorizable in the realm of real numbers.

Step 4: Discussing the Discriminant

The discriminant of a quadratic equation is \(b^{2} - 4ac\). For \(x^{2}-1\), the discriminant is \(0^{2} - 4(1)(-1) = 4\), which is a perfect square, and makes the roots real. For \(x^{2}+1\), the discriminant is \(0^{2} - 4(1)(1)= -4\), which is negative, making the roots imaginary. This demonstrates that an expression is factorable when its discriminant is a non-negative real number.