Problem 17 - Precalculus : Building Concepts and Connections

Question

Find all the zeros, real and nonreal, of the polynomial. Then express \(p(x)\) as a product of linear factors. $$p(x)=x^{2}-\sqrt{3}$$

Step-by-Step Solution

Answer

The zeros of the polynomial \(p(x)=x^{2}-\sqrt{3}\) are \(x = \sqrt[2]{\sqrt{3}}\) and \(x = -\sqrt[2]{\sqrt{3}}\) and the polynomial expressed as a product of linear factors is \(p(x) = (x - \sqrt[2]{\sqrt{3}})(x + \sqrt[2]{\sqrt{3}})\).

Step 1: Finding the Zeros

Set the polynomial equal to zero and solve for \(x\). So our equation becomes as follows: \(x^{2}-\sqrt{3} = 0\). Adding \(\sqrt{3}\) to both sides we get \(x^2 = \sqrt{3}\). We then have two possible solutions when we take the square root of \(\sqrt{3}\). Those are \(x = \sqrt[2]{\sqrt{3}}\) and \(x = -\sqrt[2]{\sqrt{3}}\). Therefore we have two roots. One is real and the other is nonreal.

Step 2: Expressing as a Product of Linear Factors

Since the roots of the polynomial are \(x = \sqrt[2]{\sqrt{3}}\) and \(x = -\sqrt[2]{\sqrt{3}}\), we can express \(p(x)\) as a product of linear factors as follows: \(p(x) = (x - \sqrt[2]{\sqrt{3}})(x + \sqrt[2]{\sqrt{3}})\).

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of degree two. It is typically written in the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In the given exercise, the polynomial you encountered is \( p(x) = x^2 - \sqrt{3} \). Here, the term \( b \) is zero, and \( c \) is \(-\sqrt{3}\). Quadratic equations can often be solved by finding the values of \( x \) that satisfy the equation set to zero. Understanding how to express a quadratic in its factored form helps reveal its roots, or solutions, which are the values of \( x \) that will make the equation equal zero. These roots are crucial for sketching graphs and understanding the behavior of polynomials.

Linear Factors

Expressing a quadratic equation as the product of linear factors means rewriting it in such a way that it breaks down into simpler expressions. The linear factors are polynomials of degree one. For the polynomial \( p(x) = x^2 - \sqrt{3} \), once we identified the zeros as \( \sqrt[2]{\sqrt{3}} \) and \( -\sqrt[2]{\sqrt{3}} \), we could express \( p(x) \) using these roots:

• \( p(x) = (x - \sqrt[2]{\sqrt{3}})(x + \sqrt[2]{\sqrt{3}}) \)

This expression shows that the polynomial can be derived by multiplying these two linear factors. Converting to linear factors provides an insight into the roots of the polynomial and is an essential part of analyzing polynomials.

Real and Nonreal Roots

Roots of a polynomial equation are points where the equation equals zero. These roots can be real or nonreal (complex). In mathematics, a real root is a solution that is a real number, while a nonreal root includes imaginary or complex numbers. For the quadratic equation in this exercise, the roots are \( x = \sqrt[2]{\sqrt{3}} \) and \( x = -\sqrt[2]{\sqrt{3}} \), both of which are real numbers. However, some quadratic equations can have nonreal roots, expressed generally in the form \( a + bi \), where \( i\) is the imaginary unit satisfying \( i^2 = -1 \). Real roots appear on the x-axis of a graph as intercepts, whereas nonreal roots do not intersect the x-axis but are relevant in complex-plane evaluations. Determining the nature—real or nonreal—of a polynomial's roots is fundamental for understanding the graph and solution of the equation.