Problem 22 - College Algebra

Question

In Exercises \(15-22,\) use the given root to find the solution set of the polynomial equation. $$ 4 x^{4}-28 x^{3}+129 x^{2}-130 x+125=0 ; 3-4 i $$

Step-by-Step Solution

Answer

The roots of the polynomial are \(3-4i\), \(3+4i\), \(1 - i\) and \(1 + i\).

Step 1: Identifying the Complex Conjugate

Find the complex conjugate of the given root. In this case, if \(3 - 4i\) is a root, then its conjugate, \(3 + 4i\), is also a root of the polynomial.

Step 2: Incorporate Roots into Polynomial

Express as an equation of the form \(4x^4 - 28x^3 + 129x^2 -130x + 125 = (x-(3-4i))(x-(3+4i))(ax^2+bx+c)\).

Step 3: Polynomial Division

Divide the given polynomial by the quadratic polynomial \(x^2 - 6x + 25\) resulting from step 2. The quotient will give the quadratic binomial \(4x^2 -4x +5\). This binomial will provide the other two real roots of the equation.

Step 4: Finding the Final Solutions

Set the quotient \(4x^2 -4x +5 = 0\). Use the quadratic formula to find the remaining roots: \(x = 1 \pm i\).

Key Concepts

Understanding Complex Conjugates

In the world of algebra, a complex number has a very special partner known as its complex conjugate. Every complex number of the form \(a + bi\) is paired with a complex conjugate \(a - bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the square root of \(-1\). This concept is particularly important when dealing with polynomial roots. For example, if a polynomial with real coefficients has a complex root like \(3 - 4i\), its complex conjugate \(3 + 4i\) is also a root. This pairing is a consequence of the complex roots theorem.

When you are given one complex root, you can instantly write down its conjugate—just flip the sign between the two terms. In the provided exercise, the given root \(3 - 4i\) comes hand in hand with the conjugate root \(3 + 4i\). Together, they form a pair of solutions to the polynomial equation, simplifying the task of finding all the roots. Complex conjugates also have a crucial role in polynomial division, where they are used to factorize polynomials into smaller degree polynomials.

Polynomial Division Simplified

Another fundamental technique in solving polynomial equations is polynomial division. It helps us to break down complex polynomials into simpler ones, which can then be solved more straightforwardly. The process is similar to long division but involves polynomials instead of numbers.

To apply polynomial division, we often start with the known roots of the polynomial. For instance, in the exercise, the known complex roots \((3 - 4i)\) and its conjugate \((3 + 4i)\) were used to create a quadratic factor \(x^2 - 6x + 25\). By dividing the original polynomial by this quadratic factor using polynomial division, you can find another polynomial of a lower degree. The result from this division (the quotient) in our exercise is the quadratic polynomial \(4x^2 - 4x + 5\), which is much easier to handle. This subsequent quadratic can be solved using various methods, most notably the quadratic formula.

The Quadratic Formula in Action

The quadratic formula is a powerful tool to find the solutions, or roots, of quadratic equations of the form \(ax^2 + bx + c = 0\). The formula itself is given by:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
It works by finding the values of x that satisfy the quadratic equation, taking into account all possible scenarios through the \(\pm\) symbol, which denotes that we can add or subtract to find two possible solutions for x. After polynomial division in our ongoing exercise, we are left with a quadratic equation \(4x^2 - 4x + 5 = 0\). By using the quadratic formula, we find the other roots of the original fourth-degree polynomial, which are \(1 + i\) and \(1 - i\). These solutions, combined with our complex conjugates from earlier, give us the complete set of roots for the polynomial equation, fully solving the problem.