Problem 60 - Algebra and Trigonometry

Question

55–64 ? 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 \(x = -3\), \(x = 3i\), and \(x = -3i\).

Step 1: Identify Possible Rational Roots

According to the Rational Root Theorem, possible rational roots of the polynomial equation \(x^3+3x^2+9x+27=0\) are factors of the constant term, \(27\). Possible factors are \(\pm 1, \pm 3, \pm 9, \pm 27\).

Step 2: Test Rational Roots with Synthetic Division

Test each possible rational root using synthetic division or substitution. Try \(-3\) as a potential root.### Synthetic Division with \(-3\)- Coefficients: \(1, 3, 9, 27\)- Synthetic Division: 1 | 3 | 9 | 27 -3 | -3 | 0 | -27 -------------------- 1 | 0 | 9 | 0 The remainder is \(0\), so \(x = -3\) is a root.

Step 3: Factor the Polynomial

Since \(x = -3\) is a root, \(x+3\) is a factor of the polynomial. Divide \(x^3+3x^2+9x+27\) by \(x+3\) to find the other factors:- The quotient from division: \(x^2 + 9\).Therefore, the factorization is \((x+3)(x^2+9)\).

Step 4: Solve the Quadratic Factor

Set the quadratic factor \(x^2 + 9 = 0\) and solve for \(x\):- \(x^2 = -9\)- \(x = \pm \sqrt{-9}\)- \(x = \pm 3i\)Thus, the complex solutions are \(x = 3i\) and \(x = -3i\).

Step 5: List All Solutions

The solutions of the equation \(x^3 + 3x^2 + 9x + 27 = 0\) are \(x = -3\), \(x = 3i\), and \(x = -3i\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a useful tool in algebra for finding possible rational solutions of polynomial equations. When dealing with a polynomial equation with integer coefficients, like the one in this exercise, the theorem helps us identify all possible rational roots quite efficiently. The roots can be viewed as potential candidates for solutions.

In simple terms, the Rational Root Theorem states that any potential rational root, given a polynomial with integer coefficients, will be a fraction \(\frac{p}{q}\), where:

• \(p\) is a factor of the constant term (the term with no variable, at the end of the polynomial),
• \(q\) is a factor of the leading coefficient (the coefficient of the highest degree term).

For the polynomial \(x^3 + 3x^2 + 9x + 27 = 0\), the constant term is 27, and its factors include \(\pm 1, \pm 3, \pm 9, \pm 27\). Since our leading coefficient in this equation is 1, any rational root must be one of these factors divided by 1, leaving them unchanged. Thus, these are the candidates you should test as possible roots.
Another helpful aspect of the Rational Root Theorem is that once you find a rational root by substituting and evaluating the polynomial, you can factor out the corresponding binomial to simplify the equation further.

Synthetic Division

Synthetic division is a streamlined method of dividing polynomials that is much less cumbersome than traditional long division. This method is particularly useful when checking potential roots identified by the Rational Root Theorem.

Consider you've found a potential root, such as \(-3\) in the equation \(x^3+3x^2+9x+27=0\). To use synthetic division:

• Write down the coefficients of the polynomial: 1, 3, 9, and 27.
• Place the potential root, \(-3\), to the left of your setup.
• Carry down the leading coefficient (1 in this case) under the division line.
• Multiply this number by \(-3\), and add this to the second coefficient.
• Repeat this multiplication and addition process for each succeeding coefficient until you reach the end.

If the remainder is zero at the end of synthetic division, then \(-3\) is indeed a root of the polynomial. For this problem, our synthetic division confirmed \(-3\) is a root, leaving us with the quotient \(x^2 + 9\). Besides verifying roots, synthetic division simplifies finding the polynomial's other factors and assists in further solving the polynomial equation.

Complex Solutions

Complex solutions arise when solving polynomial equations that lead to square roots of negative numbers. In our exercise, after applying synthetic division using \(-3\) as a root, we are left with the quadratic equation \(x^2 + 9 = 0\). Solving this introduces the concept of imaginary numbers.

An imaginary number is defined as the square root of a negative real number. In this case, solving the equation \(x^2 = -9\) gives us \(x = \pm \sqrt{-9}\). Given that \(\sqrt{-1}\) is defined as \(i\), the imaginary unit, we use it to further solve:

• First, simplify \(\sqrt{-9}\) into \(\sqrt{9} \times \sqrt{-1}\).
• This results in \(\pm 3i\), where \(i = \sqrt{-1}\).

Thus, \(x = 3i\) and \(x = -3i\) are complex solutions to the quadratic part of our original equation. It's important to note that complex solutions often come in conjugate pairs – in this case, \(3i\) and \(-3i\). When combined with our earlier real solution \(x = -3\), we get all three solutions to our original cubic equation \(x^3+3x^2+9x+27=0\). Understanding these complex solutions enriches our knowledge of how equations can be solved beyond real numbers.