Problem 54 - Algebra and Trigonometry

Question

Find all zeros of the polynomial. \(P(x)=x^{4}-2 x^{3}-2 x^{2}-2 x-3\)

Step-by-Step Solution

Answer

The zeros of the polynomial are \(x = -1, 3, i, -i\).

Step 1: Identify Potential Rational Zeros

According to the Rational Root Theorem, potential rational zeros of the polynomial \(P(x) = x^4 - 2x^3 - 2x^2 - 2x - 3\) are the factors of the constant term divided by the factors of the leading coefficient. Here, the constant term is -3 and the leading coefficient is 1. Possible rational roots are \(\pm 1, \pm 3\).

Step 2: Test Potential Zeros Using Synthetic Division or Evaluation

Evaluate the polynomial \(P(x)\) at the potential zeros: \(x = \pm 1, \pm 3\).1. \(P(1) = 1^4 - 2(1)^3 - 2(1)^2 - 2(1) - 3 = 1 - 2 - 2 - 2 - 3 = -8\).2. \(P(-1) = (-1)^4 - 2(-1)^3 - 2(-1)^2 - 2(-1) - 3 = 1 + 2 - 2 + 2 - 3 = 0\). Since \(P(-1) = 0\), \(x = -1\) is a root of the polynomial.

Step 3: Perform Polynomial Division

Divide \(P(x)=x^4-2x^3-2x^2-2x-3\) by \(x+1\) to simplify the polynomial:Use synthetic division:1. Write down the coefficients: \(1, -2, -2, -2, -3\).2. Use \(-1\) as the divisor.3. After performing the division, the quotient is \(x^3 - 3x^2 + x - 3\) with a remainder of \(0\) (confirming \(x + 1\) is a factor).

Step 4: Solve the Quotient Polynomial

To find the remaining zeros, solve the quotient polynomial \(x^3 - 3x^2 + x - 3 = 0\).Try \(x=3\) as a potential root using synthetic evaluation:1. Evaluate at \(x=3\): \(3^3 - 3(3)^2 + 3 - 3 = 27 - 27 + 3 - 3 = 0\). Since \(x = 3\) is a root, perform synthetic division on \(x^3 - 3x^2 + x - 3\) by \(x-3\) to get \(x^2 + 1\).

Step 5: Solve the Remaining Quadratic Equation

Solve the quadratic equation \(x^2 + 1 = 0\) for the remaining zeros.1. Set the equation to zero: \(x^2 + 1 = 0\).2. Solve for \(x\): \(x^2 = -1\) which gives \(x = i\) or \(x = -i\), where \(i\) is the imaginary unit.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a fundamental tool in algebra used to find the rational zeros of polynomial equations. This theorem predicts potential rational zeros based on the polynomial's coefficients. For a polynomial \( P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 \), any possible rational zero is of the form \( \frac{p}{q} \). Here:

• \( p \) is a factor of the constant term \( a_0 \).
• \( q \) is a factor of the leading coefficient \( a_n \).

For a polynomial like \( P(x) = x^4 - 2x^3 - 2x^2 - 2x - 3 \), the Rational Root Theorem helps identify possible rational zeros by considering the factors of \(-3\) and \(1\). Hence, potential rational zeros are \( \pm 1 \) and \( \pm 3 \). By using this theorem, you save time and effort as it reduces the number of values to test when searching for zeros. These candidates are then tested using synthetic division or direct substitution to ascertain if they are truly roots of the polynomial.

Synthetic Division

Synthetic division is a simplified method of dividing a polynomial by a linear binomial of the form \( x - c \). It is particularly useful because it requires fewer steps than traditional long division, especially when evaluating potential roots.To perform synthetic division, follow these steps:

• Write down only the coefficients of the terms in the polynomial.
• Place the potential root as the divisor outside the division setup.
• Bring down the leading coefficient to start the process.
• Multiply the potential root by the last number written and write the result beneath the next coefficient.
• Add the values in the column and repeat.

For example, when testing \( x = -1 \) in \( P(x) = x^4 - 2x^3 - 2x^2 - 2x - 3 \), synthetic division confirms this root, reducing the polynomial and providing a quotient of \( x^3 - 3x^2 + x - 3 \). Synthetic division thus not only verifies roots but also simplifies complex polynomials, making further root finding more manageable.

Imaginary Numbers

Imaginary numbers expand the number system so that equations like \( x^2 = -1 \) have solutions. The imaginary unit is denoted as \( i \), where \( i^2 = -1 \). When solving the polynomial's remaining quadratic part, \( x^2 + 1 = 0 \), we find:

• Rearrange to \( x^2 = -1 \).
• Solve for \( x \), resulting in \( x = i \) or \( x = -i \).

Imaginary numbers arise in equations where real solutions do not exist. They are crucial in complex number theory and have applications in engineering, physics, and advanced mathematics. Thus, recognizing the role of \( i \) extends algebraic capabilities, allowing us to solve otherwise unsolvable equations.