Problem 20 - College Algebra

Question

In Exercises \(15-22,\) use the given root to find the solution set of the polynomial equation. $$ x^{4}-x^{3}-9 x^{2}+29 x-60=0 ; 1+2 i $$

Step-by-Step Solution

Answer

The roots of the polynomial equation \(x^{4}-x^{3}-9 x^{2}+29 x-60=0\) are \(1+2i\), \(1-2i\), \(3\) and \(4\).

Step 1 Identify the Conjugate Root

Since the complex number \(1+2i\) is a root of the polynomial, then its conjugate, \(1-2i\), must also be a root.

Step 2 Formulate the Quadratic Equation

The quadratic polynomial corresponding to the two roots \(1+2i\) and \(1-2i\) is: \[(x-(1+2i))(x-(1-2i))=0\] This simplifies to: \[x^{2}-x+5=0\]

Step 3 Divide the Given Polynomial by the Quadratic

Now divide the given polynomial \(x^{4}-x^{3}-9 x^{2}+29 x-60\) by the quadratic polynomial we found: \[x^{2}-x+5\]. Doing the division, we find: \[x^{2}-x+12\]

Step 4 Find the Remaining Roots

Find the roots of the equation \(x^{2}-x+12=0\). These roots can be found by using the quadratic formula \(x = [-b±sqrt(b^{2}-4ac)]/(2a)\). Substituting \(a = 1\), \(b = -1\) and \(c = 12\) into the formula gives us two more roots: \(3\) and \(4\).

Key Concepts

Understanding Complex Roots

When dealing with polynomial equations, complex roots often come into play, especially with higher-degree polynomials. A complex root is a solution to a polynomial equation that is not a real number but involves the imaginary unit \(i\), where \(i\) is the square root of \(-1\). In our example, \(1+2i\) is a complex root of the polynomial \(x^4-x^3-9x^2+29x-60=0\).

Understanding complex roots is important because they indicate that the solutions to the polynomial are not limited to the real number line. This broadens the scope of solutions, making it essential to consider the entire set of complex numbers. Often, complex roots come in pairs known as conjugate pairs, especially in polynomials with real coefficients.

Exploring Conjugate Roots

Conjugate roots are pairs of complex numbers that have identical real parts and opposite imaginary parts. For example, if \(a+bi\) is a root, then \(a-bi\) is its conjugate root. These pairs are crucial when dealing with polynomials with real coefficients, as they always appear together.

• This means if \(1+2i\) is a root, we automatically know \(1-2i\) is also a root.
• This property significantly simplifies solving polynomials by creating symmetry that can be used strategically.

Recognizing the presence of conjugate roots allows us to form quadratic factors like \((x-(1+2i))(x-(1-2i))\). These quadratic expressions can further be simplified into real coefficients for ease of computation.

The Quadratic Formula Application

The quadratic formula is a vital tool in algebra, especially for finding the roots of quadratic equations. It is expressed as:
\[x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
In our problem, we use this formula after reducing the polynomial to find additional roots. To solve the quadratic equation \(x^2-x+12=0\) by using this formula:

• Identify \(a = 1\), \(b = -1\), and \(c = 12\).
• Substitute these values into the quadratic formula to find the solution.
• This process reveals real roots \(3\) and \(4\).

The quadratic formula is particularly useful because it provides a direct method to find roots when factoring isn’t immediately apparent.

Polynomial Factorization

Factorization is the method of writing a polynomial as a product of its factors. It is an essential concept because it breaks down complex polynomials into simpler components that are easier to solve. Factorization often begins by identifying known roots, such as complex or real ones, and using these to build factors.

In this problem, recognizing \(1+2i\) and \(1-2i\) as roots allowed us to form the quadratic factor \(x^2-x+5\).

• Next, dividing the original polynomial by this quadratic gave us \(x^2-x+12\), another factor.
• Each of these factors can reveal more about the nature of the polynomial’s roots.

By systematically applying factorization, one efficiently determines all roots of a polynomial, simplifying analysis and solution of equations drastically.