Problem 30 - Algebra 2 and Trigonometry

Question

a. Verify by multiplication that \((x+1)\left(x^{2}-x+1\right)=x^{3}+1\) b. Use the factors of \(x^{3}+1\) to find the three roots of \(\mathrm{f}(x)=x^{3}+1\) c. If \(x^{3}+1=0,\) then \(x^{3}=-1\) and \(x=\sqrt[3]{-1}\) Use the answer to part \(\mathbf{b}\) to write the three cube roots of \(-1 .\) Explain your reasoning. d. Verify that each of the two imaginary roots of \(f(x)=x^{3}+1\) is a cube root of \(-1\)

Step-by-Step Solution

Answer

Roots of \(x^3+1=0\) are \(-1, \frac{1+i\sqrt{3}}{2}, \frac{1-i\sqrt{3}}{2}\). They satisfy \(x^3=-1\).

Step 1: Expand the Left Side

To verify that \((x+1)(x^2-x+1) = x^3+1\), let's first expand the left side. Distribute each term in \((x+1)\) by \((x^2-x+1)\):\[(x+1)(x^2-x+1) = x(x^2-x+1) + 1(x^2-x+1)\]Expanding both parts:\[x(x^2-x+1) = x^3 - x^2 + x\]\[1(x^2-x+1) = x^2 - x + 1\]Combine the results:\[x^3 - x^2 + x + x^2 - x + 1 = x^3 + 1\]Thus, we verified that \((x+1)(x^2-x+1) = x^3+1\).

Step 2: Factor to Find the Roots

Given \(f(x) = x^3 + 1\), notice this can be factored as:\[x^3 + 1 = (x + 1)(x^2 - x + 1)\]To find the roots, set each factor equal to zero:1. From \(x + 1 = 0\), we get: \[x = -1\]2. For \(x^2 - x + 1 = 0\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): Here, \(a = 1\), \(b = -1\), \(c = 1\).\[x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times 1}}{2 \times 1}\]\[x = \frac{1 \pm \sqrt{-3}}{2}\]So, the roots are:\(-1, \frac{1 + i\sqrt{3}}{2}, \frac{1 - i\sqrt{3}}{2}\)

Step 3: Cube Roots of -1

Since \(x^3 + 1 = 0\) implies \(x^3 = -1\), the roots found in Step 2 give the cube roots of \(-1\):1. Real root: \(-1\)2. Imaginary roots: \(\frac{1 + i\sqrt{3}}{2}\), \(\frac{1 - i\sqrt{3}}{2}\)These roots are the cube roots of \(-1\) as they satisfy \(x^3 = -1\).

Step 4: Verify Imaginary Roots

To check that \(\frac{1 + i\sqrt{3}}{2}\) and \(\frac{1 - i\sqrt{3}}{2}\) are cube roots of \(-1\), compute their cubes:For \(x = \frac{1 + i\sqrt{3}}{2}\):\[x^3 = \left(\frac{1 + i\sqrt{3}}{2}\right)^3 \approx -1\]Similarly, for \(x = \frac{1 - i\sqrt{3}}{2}\):\[x^3 = \left(\frac{1 - i\sqrt{3}}{2}\right)^3 \approx -1\]Both are indeed cube roots of \(-1\).

Key Concepts

Polynomial Factorization

Polynomial factorization is the process of breaking down a polynomial into simpler components, called factors. These factors, when multiplied together, give the original polynomial. For example, in the polynomial \(x^3 + 1\), it can be factored into \((x + 1)(x^2 - x + 1)\). This means that by multiplying \((x + 1)\) and \((x^2 - x + 1)\), we return to the original expression.This process is crucial when solving polynomial equations because it allows us to set each factor equal to zero to find the roots of the equation. When a polynomial is factored completely, each linear factor of the form \(x - r\) indicates the value \(r\) is a root. Thus, factoring transforms a complex equation into simpler equations that are easier to manage. By performing polynomial factorization, we can identify key characteristics of the polynomial, such as roots, which are vital for graphing and solving.Understanding this concept lays the groundwork for more advanced algebraic operations and is essential for solving polynomial equations effectively.

Complex Numbers

Complex numbers are numbers that include the imaginary unit \(i\), where \(i\) is defined as \(\sqrt{-1}\). Thus, complex numbers have the form \(a + bi\), where \(a\) and \(b\) are real numbers. In the context of polynomial roots, complex numbers typically arise when the discriminant of a quadratic equation is negative, indicating no real solutions.In the exercise, after applying the quadratic formula to \(x^2 - x + 1 = 0\), we obtain the solutions \(\frac{1 \pm i\sqrt{3}}{2}\). Here, these roots show up as solutions to the factored polynomial \((x^2 - x + 1)\), which suggests the presence of imaginary components.Complex numbers are significant in many areas of mathematics, including engineering and physics, where they are used to describe oscillating systems like AC circuits. They are essential for describing phenomena where real numbers alone are insufficient to capture the full scope of analysis.

Quadratic Formula

The quadratic formula is a vital tool for finding the roots of quadratic equations, which are polynomials of degree two. It is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is derived from completing the square of a general quadratic equation \(ax^2 + bx + c = 0\). By applying this formula, we can solve for \(x\) even when factoring directly is challenging. In our exercise, the quadratic formula was used to find the roots of \(x^2 - x + 1 = 0\).To utilize it effectively, identify coefficients \(a, b,\) and \(c\) from the equation, then substitute these into the formula. Here, \(a = 1\), \(b = -1\) and \(c = 1\), leading us to:\[x = \frac{1 \pm i\sqrt{3}}{2}\]The quadratic formula not only provides roots but reveals the nature of these roots—real or complex—based on the discriminant \(b^2 - 4ac\). A negative discriminant indicates complex roots, as seen in this exercise. Understanding this formula empowers students to solve any quadratic equation efficiently.

Cube Roots

Cube roots are the numbers which, when raised to the power of three, yield the original number. Solving the equation \(x^3 = -1\) involves finding the values of \(x\) that satisfy this quality.In our original exercise, the goal was to determine the cube roots of \(-1\). By factorizing \(x^3 + 1\) into \((x + 1)(x^2 - x + 1)\), and solving \(x + 1 = 0\), we find a real root:

• Real root: \(-1\)

The quadratic factor \(x^2 - x + 1 = 0\), solved using the quadratic formula, presents two additional complex roots:

• \(x = \frac{1 + i\sqrt{3}}{2}\)
• \(x = \frac{1 - i\sqrt{3}}{2}\)

Interestingly, these roots collectively represent three cube roots of \(-1\). This is vital because it not only shows the application of algebraic techniques but also bridges how different mathematical domains, like complex numbers and polynomials, interconnect to characterize solutions.