Problem 16 - Basic Technical Mathematics with Calculus

Question

Find the remaining roots of the given equations using synthetic division, given the roots indicated. $$4 n^{4}+28 n^{3}+61 n^{2}+42 n+9=0 \quad(-3\text { is a double root) }$$

Step-by-Step Solution

Answer

The remaining roots are \(n = -\frac{1}{2}\) (repeated root) and \(-3\) (double root).

Step 1: Write Down the Polynomial and Known Root

The polynomial given is \(4n^4 + 28n^3 + 61n^2 + 42n + 9\). It is provided that \(-3\) is a double root, meaning \((n + 3)^2\) is a factor of the polynomial.

Step 2: Perform Synthetic Division (First Round)

Perform synthetic division using \(-3\) as a root on the polynomial. Write the coefficients [4, 28, 61, 42, 9]. Bring down the 4. Multiply \(-3\) by 4, add to the next coefficient, and continue this process: \(4, 16, 13, 3, 0\). The remainder is 0, confirming \(n + 3\) is a factor.

Step 3: Perform Synthetic Division (Second Round)

Using the result \(4n^3 + 16n^2 + 13n + 3\) from the first division, perform synthetic division with \(-3\) again. Follow the same steps: the coefficients start as [4, 16, 13, 3]. After division, the output is \(4, 4, 1, 0\). The final result is \(4n^2 + 4n + 1\). The remainder is 0, confirming \(n + 3\) was indeed a double root.

Step 4: Solve the Quadratic Polynomial

Solve the quadratic equation \(4n^2 + 4n + 1 = 0\) using the quadratic formula: \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 4\), \(b = 4\), \(c = 1\). Calculate the discriminant: \(b^2 - 4ac = 16 - 16 = 0\). Therefore, \(n = \frac{-4}{8} = -\frac{1}{2}\). This root is a repeated root since the discriminant is 0.

Key Concepts

Polynomial Roots

To grasp the concept of polynomial roots, we need to understand what happens when a polynomial is set equal to zero. The solutions to the equation are called the polynomial's roots. In the given problem, the polynomial is of fourth degree: \[4n^4 + 28n^3 + 61n^2 + 42n + 9 = 0\].

• A fourth-degree polynomial can have up to four roots.
• Each root represents a value of \(n\) that satisfies the equation.
• If a polynomial has a factor like \((n + c)\), then \(n = -c\) is a root of the polynomial.

Given that \(-3\) appears as a double root, you would apply synthetic division twice with \(-3\) to simplify the polynomial further. This verifies that \((-3)\) indeed satisfies the equation, confirming the factor has roots as specified. Identifying all roots involves using these techniques and may often require a combination of synthetic division and solving smaller resulting polynomials.

Quadratic Formula

As polynomials are reduced in complexity, you'll often get terms that can be managed through the quadratic formula. Once we've reduced the polynomial using synthetic division twice, the equation simplifies to a quadratic form: \[4n^2 + 4n + 1 = 0\].
This is where the quadratic formula becomes useful:

• The formula is \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• It's applicable when you have a polynomial of the form \(ax^2 + bx + c = 0\).

For our example, inserting values, \(a = 4\), \(b = 4\), \(c = 1\), the discriminant calculates as:\[b^2 - 4ac = 16 - 16 = 0\].
The discriminant is key:

• If greater than zero, two distinct real roots exist.
• If zero, exactly one repeated real root.
• If less than zero, no real roots exist, only complex.

In this polynomial, the discriminant is zero, giving \(n = -\frac{1}{2}\) as a repeated root.

Repeated Roots

Understanding repeated roots is essential for deciphering polynomial characteristics. A root is repeated if it appears more than once as a solution to the polynomial equation. In the problem, we encounter multiple instances of such roots:

• Initially, we know \(-3\) is a double root.
• This means \((n+3)^2\) is a factor of the polynomial \(4n^4 + 28n^3 + 61n^2 + 42n + 9\).

Repeated roots often arise from the behavior of polynomials as they "flatten" at certain values. After removing the impact of \(-3\) using synthetic division twice, we identify another repeated root using the quadratic formula for \(n = -\frac{1}{2}\):

• Repeated roots lead to the polynomial having a "touching" effect on the x-axis at that root.
• They carry implications for the graph and stability of the polynomial's expression.

Recognizing these allows us to completely factor, solve, and understand the original polynomial's behavior and solutions.