Problem 61 - Precalculus Mathematics for Calculus

Question

Find all zeros of the polynomial. $$P(x)=x^{5}-3 x^{4}+12 x^{3}-28 x^{2}+27 x-9$$

Step-by-Step Solution

Answer

The zeros are 1, 3, -3, \( i\sqrt{7} \), and \( -i\sqrt{7} \).

Step 1: Apply the Rational Root Theorem

The Rational Root Theorem suggests that if \( P(x) \) has a rational zero \( p/q \), then \( p \) is a factor of the constant term (-9), and \( q \) is a factor of the leading coefficient (1). For our polynomial, this means the possible rational roots are ±1, ±3, ±9.

Step 2: Test Possible Rational Roots

Evaluate \( P(x) \) using these candidates. By substituting these values, we find that \( P(1) = 0 \), so \( x = 1 \) is a root.

Step 3: Division of Polynomial using Synthetic Division

Since \( x = 1 \) is a root, perform synthetic division of \( P(x) \) by \( x - 1 \). This reduces the polynomial to a quotient of \( x^4 - 2x^3 + 10x^2 - 18x + 9 \).

Step 4: Repeat Rational Root Theorem on Reduced Polynomial

Apply the Rational Root Theorem again to \( x^4 - 2x^3 + 10x^2 - 18x + 9 \), with possible rational roots: ±1, ±3, ±9. Evaluating these, we find \( x = 3 \) is a root.

Step 5: Further Polynomial Division

Perform synthetic division of the reduced polynomial \( x^4 - 2x^3 + 10x^2 - 18x + 9 \) by \( x - 3 \) to obtain \( x^3 + x^2 + 7x + 3 \).

Step 6: Solve Remaining Polynomial

Use the Rational Root Theorem and synthetic division for \( x^3 + x^2 + 7x + 3 \). Testing possible roots, we find \( x = -3 \) is a root.

Step 7: Final Polynomial Division

Perform synthetic division of \( x^3 + x^2 + 7x + 3 \) by \( x + 3 \) to reduce it to \( x^2 + 7 \).

Step 8: Solve Quadratic Equation

Solve the equation \( x^2 + 7 = 0 \), which has complex solutions: \( x = i ext{√}{7} \) and \( x = -i ext{√}{7} \).

Step 9: List All Zeros of the Original Polynomial

The zeros of \( P(x) = x^5 - 3x^4 + 12x^3 - 28x^2 + 27x - 9 \) are 1, 3, -3, \( i ext{√}{7} \), and \( -i ext{√}{7} \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool used in algebra to identify potential rational zeros of a polynomial. It states that if a polynomial function \( P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \) has any rational zeros, they must be in the form of \( \frac{p}{q} \), where \( p \) is a factor of the constant term \( a_0 \) and \( q \) is a factor of the leading coefficient \( a_n \).

For example, in the polynomial \( P(x)=x^5-3x^4+12x^3-28x^2+27x-9 \), the constant term is -9, and the leading coefficient is 1. This means the possible rational roots are ±1, ±3, and ±9, because these are the factors of -9, and the only factor of 1 is 1 itself.

• Make a list of all potential rational roots using the theorem.
• Test each candidate by substituting it into the polynomial to see if it resolves to zero.

This theorem narrows down the roots we need to test, simplifying the process of factoring a polynomial.

Synthetic Division

Synthetic division is a simpler alternative to long division for polynomials, especially useful when dealing with linear divisors of the form \( x - c \). It streamlines the process of dividing a polynomial by taking advantage of the roots found using the Rational Root Theorem.

To perform synthetic division:

• Write down the coefficients of the polynomial.
• Place the root (e.g., \( c \)) outside the division bracket.
• Bring down the leading coefficient.
• Multiply the root by this leading coefficient, and add to the next coefficient.
• Repeat the process for the entire row of coefficients.

In our problem, once it was determined that \( x = 1 \) is a root, synthetic division was used with \( 1 \) to reduce the polynomial from degree 5 to degree 4. This simplified polynomial can then be further divided in the same manner.

Complex Numbers

Complex numbers extend the idea of numbers beyond the real number system by including an imaginary unit, denoted as \( i \), where \( i^2 = -1 \). This allows for the existence of solutions to equations that have no real roots.

In the given polynomial, the reduced quadratic \( x^2 + 7 = 0 \) has no real number solutions, since solving it gives nth ode. Specifically, the solutions are \( x = i\sqrt{7} \) and \( x = -i\sqrt{7} \), where \( \sqrt{7} \) is a real number multiplied by the imaginary unit \( i \).

• Complex solutions often appear as conjugate pairs when solving polynomials with real coefficients.
• These numbers form the basis for complex plane mathematics, going beyond the simple line of real numbers.

Understanding complex numbers is key to solving polynomial equations that do not factor easily into real numbers.

Quadratic Equation

A quadratic equation is a second-degree polynomial in the form \( ax^2 + bx + c = 0 \). It is characterized by having at most two solutions, which can be real or complex.

When solving quadratic equations, several methods can be used:

• Factoring, when the quadratic easily breaks down into two binomials.
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), particularly when other methods are not straightforward.
• Completing the square, useful for deriving the quadratic formula or specific scenarios.

In our problem, the final step involved solving the quadratic equation \( x^2 + 7 = 0 \). Since the equation cannot be factored over the reals, the quadratic equation's solutions are complex, incorporating the concept of imaginary numbers \( i \), as demonstrated by the solutions \( i\sqrt{7} \) and \( -i\sqrt{7} \). This showcases the versatility and scope of quadratics beyond just real-number solutions.