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Chapter 6: Problem 113

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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.
04

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.

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Most popular questions from this chapter

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$-4 y^{3}+28 y^{2}-40 y$$

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Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$9 y^{2}+64$$

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Contain polynomials in several variables. Factor each polynomial completely and check using multiplication. $$3 x^{4} y^{2}-3 x^{2} y^{2}$$
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