Problem 14 - Precalculus

Question

An equation is given, followed by one or more roots of the equation. In each case, determine the remaining roots. $$\begin{array}{l}x^{5}-5 x^{4}+30 x^{3}+18 x^{2}+92 x-136=0 \\\x=-1+i \sqrt{3}, x=3-5 i\end{array}$$

Step-by-Step Solution

Answer

The remaining root is \( x = -1 \).

Step 1: Understanding complex roots

The given equation is a polynomial of degree 5, which means it will have 5 roots. Complex roots of polynomials with real coefficients always occur in conjugate pairs. The given roots are \( x = -1 + i\sqrt{3} \) and \( x = 3 - 5i \). Therefore, their conjugates \( x = -1 - i\sqrt{3} \) and \( x = 3 + 5i \) are also roots of the equation.

Step 2: Factoring polynomial using known roots

Since we have four roots \( x = -1 + i\sqrt{3}, x = -1 - i\sqrt{3}, x = 3 - 5i, \text{ and } x = 3 + 5i \), we can form quadratic factors from these pairs. The factors corresponding to the roots \( x = -1 + i\sqrt{3} \) and \( x = -1 - i\sqrt{3} \) are \[ (x - (-1 + i\sqrt{3}))(x - (-1 - i\sqrt{3})) = (x + 1)^2 + 3. \] Similarly, the factors for the roots \( x = 3 - 5i \) and \( x = 3 + 5i \) are \[ (x - (3 - 5i))(x - (3 + 5i)) = (x - 3)^2 + 25. \]

Step 3: Determine the remaining root

Multiply the quadratic factors obtained from step 2: \[ (x + 1)^2 + 3 = x^2 + 2x + 4 \] and \[ (x - 3)^2 + 25 = x^2 - 6x + 34. \] Multiply these two resulting quadratic polynomials: \[ (x^2 + 2x + 4)(x^2 - 6x + 34) = x^4 - 4x^3 + 44x^2 - 196x + 136. \] Since this product gives us a degree 4 polynomial, and our original polynomial is degree 5, the missing factor should be linear. Divide the original polynomial \( x^5 - 5x^4 + 30x^3 + 18x^2 + 92x - 136 \) by \( x^4 - 4x^3 + 44x^2 - 196x + 136 \) to find the missing linear factor. The result is \( x + 1 \), indicating that \( x = -1 \) is the missing root.

Key Concepts

Degree of Polynomial

To start, it's essential to understand what the degree of a polynomial is. The degree is the highest power of the variable in a polynomial expression. It tells us how many roots, or solutions, the polynomial will have.
For instance, in this exercise, we are dealing with a polynomial of degree 5, which means the equation has 5 roots in total. This includes real and complex roots. The importance of the degree cannot be overstated as it not only tells us how many roots to expect but also guides us in solving the equation.
The degree helps us determine how complicated the polynomial might be and forms the basis for using theorems and solving strategies. For instance, in solving our given polynomial equation, determining the degree lets us expect five roots to find. This understanding is pivotal in tackling polynomial equations effectively.

Complex Conjugates

When dealing with polynomials with real coefficients, complex roots often appear as conjugate pairs. This means if \(a+bi\) is a root, then \(a-bi\) must also be a root. This property stems from the necessity for non-real roots to maintain real coefficients in a polynomial.
This exercise provides two complex roots, \(-1 + i\sqrt{3}\) and \(3 - 5i\). Following the rule of conjugates, their partners \(-1 - i\sqrt{3}\) and \(3 + 5i\) also become roots of the polynomial. It is always helpful to remember this property while solving for polynomial roots to accurately find all roots.
Utilizing this conjugate relationship simplifies finding the polynomial factors, as it automatically provides additional roots when a complex root is identified. This ability to identify and apply complex conjugates is a powerful tool in algebra when handling polynomials with real coefficients.

Factoring Polynomials

Factoring polynomials is indispensable in finding roots or solving polynomial equations. In this context, once you establish some roots—real or complex—you can form quadratic or linear factors from these roots.
For the provided polynomial, we use the complex conjugate pairs to form quadratic factors:

• \((x + 1)^2 + 3\) comes from \(-1 \pm i\sqrt{3}\).
• \((x - 3)^2 + 25\) comes from \(3 \pm 5i\).

Multiplying these gives a degree 4 polynomial, reflecting four roots. Since the original polynomial is degree 5, the remaining root must come from a linear factor.
This linear factor \(x + 1\) indicates there is a repeated root at \(x = -1\). Correctly understanding and applying factoring helps decompose polynomials effectively and find all roots systematically, especially when handled alongside complex conjugates and polynomial degrees. Understanding this concept thoroughly enhances problem-solving strategies in algebra.