Problem 7 - Precalculus

Question

An equation is given, followed by one or more roots of the equation. In each case, determine the remaining roots. $$4 x^{3}-47 x^{2}+232 x+61=0 ; x=6-5 i$$

Step-by-Step Solution

Answer

The remaining root is \(x = \frac{1}{4}\).

Step 1: Identify the Conjugate Root

Since the coefficients of the polynomial are real numbers, complex roots must occur in conjugate pairs. Given that one root is \(x = 6 - 5i\), the conjugate root also must be \(x = 6 + 5i\).

Step 2: Set Up the Quadratic Factor

The conjugate roots \(x = 6 - 5i\) and \(x = 6 + 5i\) suggest that \((x - (6 - 5i))(x - (6 + 5i))\) is a factor of the polynomial. We can express this as \((x - 6 + 5i)(x - 6 - 5i)\).

Step 3: Expand the Quadratic Factor

Use the difference of squares formula to factor: \[(x - 6)^2 - (5i)^2 \]This simplifies to:\[(x - 6)^2 + 25\]Expanding \((x - 6)^2\):\[x^2 - 12x + 36\]So, the expression becomes:\[x^2 - 12x + 36 + 25 = x^2 - 12x + 61\]

Step 4: Divide the Original Polynomial by the Quadratic Factor

Use polynomial division or synthetic division to divide the original cubic polynomial, \(4x^3 - 47x^2 + 232x + 61\), by \(x^2 - 12x + 61\). This will give the remaining factor, which is linear.

Step 5: Finding the Remaining Root

Upon dividing, we find that the quotient is \(4x - 1\), which we set equal to zero to find the remaining root:\[4x - 1 = 0\]Solving for \(x\):\[x = \frac{1}{4}\]

Key Concepts

Complex Conjugates

When dealing with polynomials that have real coefficients, it's essential to understand the concept of complex conjugates. These are pairs of complex numbers that have identical real parts but opposite imaginary parts. For instance, if one root of the polynomial is \(6 - 5i\), its complex conjugate will be \(6 + 5i\). This phenomenon occurs because non-real roots in polynomials with real coefficients always come in conjugate pairs.

This property ensures that any imaginary parts in a quadratic factor cancel out, leaving a polynomial with real coefficients. Conjugate pairs are vital in preserving the integrity of real-valued polynomials, allowing us to factor and simplify equations efficiently. Remember:

• Complex roots appear in conjugate pairs when coefficients are real.
• The conjugate of \(a + bi\) is \(a - bi\).

Understanding conjugates simplifies the process of finding additional roots and factors in polynomial equations.

Polynomial Division

Polynomial division is a powerful method to break down a polynomial into simpler, more manageable components. In this exercise, after establishing the quadratic factor using conjugate roots, the task involves dividing the original polynomial by this factor to uncover remaining roots.

This is akin to dividing numbers but involves variables. There are various methods, including long division and synthetic division, to achieve this. Typically:

• Set up the division with the original polynomial as the dividend and the known factor as the divisor.
• Use division rules to simplify and reduce the polynomial step by step until you obtain a linear quotient (if possible).

This linear quotient often contains the remaining roots as its solution, as seen in this exercise where division revealed the linear quotient \(4x - 1\). Once you solve this, you'll find the remaining root of the polynomial.

Quadratic Factors

Quadratic factors are key components of polynomial equations, especially when dealing with complex roots. They are expressions of the form \(ax^2 + bx + c\) that multiply to give the original polynomial.

In this exercise, the complex roots \(6 - 5i\) and \(6 + 5i\) form the quadratic factor \((x - 6)^2 + 25\) or \(x^2 - 12x + 61\) after simplification using the difference of squares formula.

This is a crucial step because it transforms complex components into a real-number polynomial, making it easier to handle. Quadratic factors:

• Encapsulate pairs of conjugate complex roots.
• Allow the entire polynomial to be expressed in terms of simpler parts.
• Provide a clear path for using polynomial division to find other roots.

These factors enable further simplification and problem-solving in polynomial equations, forming an essential part of algebraic operations.