Problem 66 - Algebra and Trigonometry

Question

\(61-70\) . Find all solutions, real and complex, of the equation. $$ x^{3}+3 x^{2}+9 x+27=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = -3, 3i, -3i\).

Step 1: Check for Factorability

The first step is to test if the polynomial can be factorized easily. We are looking for a root of the polynomial by checking simple values like \(x = -1, 0, 1, -2, 2, \) etc. Let's try \(x = -3\). We substitute \(x = -3\) into the polynomial: \((-3)^3 + 3(-3)^2 + 9(-3) + 27\). This simplifies to: \(-27 + 27 - 27 + 27 = 0\). Hence \(x = -3\) is a root.

Step 2: Use Synthetic Division

Perform synthetic division on the polynomial using \(x = -3\) which we confirmed as a root. The coefficients are 1, 3, 9, 27. The process of synthetic division goes as follows: 1. Write coefficient: 1 2. Multiply by root (-3): -3 3. Add to next coefficient (3): 0 4. Multiply by root (-3): 0 5. Add to next coefficient (9): 9 6. Multiply by root (-3): -27 7. Add to next coefficient (27): 0The quotient from the division is \(x^2 + 0x + 9\) or simply \(x^2 + 9\). The remainder is 0, which confirms the division is exact.

Step 3: Solve the Quadratic Equation

Now, solve the quadratic equation \(x^2 + 9 = 0\). This can be rewritten as \(x^2 = -9\). Taking the square root of both sides gives \(x = \pm \sqrt{-9}\). Simplifying further using imaginary numbers, we have \(x = \pm 3i\). Thus, the solutions from the quadratic are complex numbers \(x = 3i\) and \(x = -3i\).

Step 4: List All Solutions

We now list all solutions found. In Step 1, we found a real root \(x = -3\). In Step 3, we found two complex solutions, \(x = 3i\) and \(x = -3i\). Therefore, the complete set of solutions (real and complex) for the given polynomial equation \(x^3 + 3x^2 + 9x + 27 = 0\) are \(x = -3, 3i, -3i\).

Key Concepts

Synthetic Division

Synthetic division is a quick method for dividing a polynomial by a linear binomial of the form \(x - c\). It's often used to simplify polynomial division and confirm roots of the polynomial equation.
To use synthetic division, start by writing down the coefficients of the polynomial. Then, write the value of the root you are testing outside the division bracket. For instance, in our problem, we are dividing the polynomial \(x^3 + 3x^2 + 9x + 27\) by \(x + 3\) because \(x = -3\) is a known root.
Follow these steps:

• Write the coefficients: 1, 3, 9, 27.
• Place \(-3\) outside the division bracket.
• Bring down the leading coefficient (1) as is.
• Multiply \(-3\) by this leading coefficient to get \(-3\).
• Add this product to the next coefficient (3) to get 0.
• Repeat this multiply-and-add process down the line.

The final row represents the coefficients of the quotient polynomial. Since the remainder is 0, \(x + 3\) is indeed a factor. The quotient is \(x^2 + 9\).
Synthetic division offers a neat way to strip down the polynomial, making the next steps easier. It saves time compared to long division, especially with higher-degree polynomials.

Complex Numbers

Complex numbers are numbers that include a real and an imaginary part and are usually expressed in the form \(a + bi\). Imaginary numbers arise naturally when dealing with square roots of negative numbers, such as \(i\), which is the square root of \(-1\). In our exercise, we encounter complex solutions when solving the quadratic equation \(x^2 + 9 = 0\).
To solve this equation, note:

• Get \(x^2 = -9\) by isolating \(x\).
• Taking the square root of both sides results in \(x = \pm \sqrt{-9}\).
• This simplifies to \(x = \pm 3i\), using \(i\) for the imaginary unit.

These solutions, \(x = 3i\) and \(x = -3i\), demonstrate the use of complex numbers to provide a full set of solutions.
Complex numbers are crucial in algebra, especially in problems like this where real numbers alone aren't sufficient. They extend our capability to solve equations beyond the real number limits, allowing us to understand the complete behavior of polynomial functions.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, generally in the form \(ax^2 + bx + c = 0\). Solving these is a critical skill, often leveraging techniques such as factoring, using the quadratic formula, or completing the square.
In our solved exercise, after using synthetic division, we arrived at the quadratic \(x^2 + 9 = 0\). Here's how you solve it:

• Recognize it's in standard form with \(a=1, b=0, c=9\).
• Since it can't be factored using real numbers, consider taking the square root on both sides.
• This gives \(x^2 = -9\), leading to \(x = \pm \, 3i\) when solved for \(x\).

Quadratic equations appear across many aspects of math and science due to their simplicity and power. They are foundational for understanding higher-level polynomial solutions and naturally lead us to complex numbers when solutions in the real number set are unavailable.
Understanding how to solve them, whether through factoring, the quadratic formula, or by recognizing equations that lead to complex numbers, is crucial for success in algebra and calculus.