Problem 48 - Algebra and Trigonometry

Question

\(41-58=\) Find all zeros of the polynomial. $$ P(x)=2 x^{3}-8 x^{2}+9 x-9 $$

Step-by-Step Solution

Answer

The zeros are \( x = 3, \frac{1+\sqrt{5}i}{2}, \frac{1-\sqrt{5}i}{2} \).

Step 1: Understanding the Problem

We need to find all the zeros of the polynomial \( P(x)=2x^3-8x^2+9x-9 \). A zero of a polynomial is a value of \( x \) for which \( P(x)=0 \).

Step 2: Check for Rational Roots

By the Rational Root Theorem, the potential rational roots of \( P(x) \) are factors of the constant term \(-9\), divided by factors of the leading coefficient \(2\). The potential rational roots are \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2} \).

Step 3: Using Synthetic Division to Test Roots

Let's test \( x = 3 \). Using synthetic division:1. Coefficients are \( 2, -8, 9, -9 \).2. Bring down the 2.3. Multiply by 3 and add: \( 2 \to (2 \times 3)-8 = -2 \).4. Multiply by 3 and add: \( -2 \to (-2 \times 3)+9 = 3 \).5. Multiply by 3 and add: \( 3 \to (3 \times 3)-9 = 0 \).Thus, \( x = 3 \) is a root.

Step 4: Factor the Polynomial

Since \( x = 3 \) is a root, \( x - 3 \) is a factor of \( P(x) \). After synthetic division, the quotient is \( 2x^2 - 2x + 3 \). Therefore, \( P(x) = (x - 3)(2x^2 - 2x + 3) \).

Step 5: Find Remaining Roots

Set \( 2x^2 - 2x + 3 = 0 \) and use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 2, b = -2, c = 3 \).Calculate the discriminant: \( b^2 - 4ac = (-2)^2 - 4 \cdot 2 \cdot 3 = 4 - 24 = -20 \), which indicates complex roots.

Step 6: Calculate Complex Roots

Since the discriminant is \(-20\), the roots are complex:\[ x = \frac{-(-2) \pm \sqrt{-20}}{2 \cdot 2} = \frac{2 \pm \sqrt{20}i}{4} = \frac{2 \pm 2\sqrt{5}i}{4} = \frac{1 \pm \sqrt{5}i}{2} \]

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a handy tool for finding possible rational zeros of a polynomial. It states that any rational solution, or root, of a polynomial equation with integer coefficients, is a fraction \( \frac{p}{q} \), where **p** is a factor of the constant term and **q** is a factor of the leading coefficient. This theorem helps us narrow down the list of values to test when looking for roots. In the example polynomial \( P(x)=2x^3-8x^2+9x-9 \), we use this theorem to find that the potential rational roots are:

• \( \pm 1, \pm 3, \pm 9 \)
• \( \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2} \)

These possible roots are based on the factors of the constant term \(-9\) and the leading coefficient \(2\). This theorem dramatically reduces the number of possible roots to check.

Synthetic Division

Synthetic division is a simplified method of dividing polynomials, especially useful when evaluating a polynomial at a given value or when checking potential roots. It is more efficient than traditional long division for polynomials, particularly when the divisor is of the form \( x - c \). Here is how it works:- Write down only the coefficients of the polynomial.- Bring down the leading coefficient.- Multiply it by the proposed root, then add it to the next coefficient.- Repeat this process until the last number.In our example, testing \( x = 3 \) as a root, we start with the coefficients \( 2, -8, 9, -9 \). After performing the synthetic division, we verify that \( x = 3 \) is indeed a root because the remainder is zero. This confirmation allows us to factor \( (x - 3) \) from the polynomial.

Quadratic Formula

When a polynomial has been simplified to a quadratic form, the quadratic formula can be used to find its roots. The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula calculates the roots for the quadratic equation \( ax^2 + bx + c = 0 \). It's especially useful when the equation doesn't factor easily. In the reduced polynomial from our exercise, \( 2x^2 - 2x + 3 = 0 \), we apply the quadratic formula. Here, \( a = 2 \), \( b = -2 \), and \( c = 3 \). The discriminant, \( b^2 - 4ac \), is calculated as \(-20\), which is negative, indicating complex roots.

Complex Roots

Complex roots occur when the discriminant in the quadratic formula is negative, meaning the square root of a negative number is needed. This introduces imaginary numbers, represented by \( i \), where \( i = \sqrt{-1} \). In our exercise, the discriminant of \( 2x^2 - 2x + 3 \) was \(-20\), so the complex roots are found using:\[ x = \frac{2 \pm \sqrt{-20}}{4} = \frac{2 \pm 2\sqrt{5}i}{4} = \frac{1 \pm \sqrt{5}i}{2} \]These solutions, \( \frac{1 + \sqrt{5}i}{2} \) and \( \frac{1 - \sqrt{5}i}{2} \), are the complex roots of the quadratic part of the polynomial. They reveal that not all zeros of a polynomial need to be real numbers. Embracing complex roots expands the scope of solutions to any polynomial equation.