Problem 51 - Precalculus Mathematics for Calculus

Question

Find all zeros of the polynomial. $$P(x)=2 x^{3}+7 x^{2}+12 x+9$$

Step-by-Step Solution

Answer

The zeros of the polynomial are \( x = -1, \frac{-5 + i \sqrt{47}}{4}, \frac{-5 - i \sqrt{47}}{4} \).

Step 1: Check for Rational Zeros

First, list all rational factors of the constant term (9) and the leading coefficient (2). The possible rational zeros are \( \pm 1, \pm 3, \pm 9 \) and each over \( \pm 1, \pm 2 \). Evaluate these potential zeros using the polynomial function \( P(x) = 2x^3 + 7x^2 + 12x + 9 \) to narrow down candidates.

Step 2: Use Synthetic Division to Verify Zeros

Check the potential zeros from Step 1 using synthetic division. For example, try \( x = -1 \). Performing synthetic division on the coefficients [2, 7, 12, 9] with \( x = -1 \) results in zero, confirming \( x = -1 \) is a zero of \( P(x) \). This produces a quotient: \( 2x^2 + 5x + 9 \).

Step 3: Solve the Quadratic Equation

Solve the remaining quadratic equation \( 2x^2 + 5x + 9 = 0 \) using the quadratic formula. The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) gives \( a = 2, b = 5, \) and \( c = 9 \). Calculate the discriminant: \( b^2 - 4ac = 5^2 - 4(2)(9) = -47 \). Since the discriminant is negative, there are two complex zeros.

Step 4: Compute Complex Zeros

Calculate the complex zeros: \( x = \frac{-5 \pm \sqrt{-47}}{4} = \frac{-5 \pm i \sqrt{47}}{4} \). The complex zeros are \( x = \frac{-5 + i \sqrt{47}}{4} \) and \( x = \frac{-5 - i \sqrt{47}}{4} \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a handy tool for finding potential rational zeros of a polynomial. It tells us that if a polynomial has a rational zero, it must be of the form \( \frac{p}{q} \), where \( p \) is a factor of the constant term, and \( q \) is a factor of the leading coefficient.
For the polynomial \( P(x) = 2x^3 + 7x^2 + 12x + 9 \), the constant term is 9 and the leading coefficient is 2. This means that potential rational zeros come from the factors of 9 (\( \pm 1, \pm 3, \pm 9\)) over the factors of 2 (\( \pm 1, \pm 2\)).
Listing these, we get possible zeros: \( \pm 1, \pm \frac{1}{2}, \pm 3, \pm \frac{3}{2}, \pm 9, \pm \frac{9}{2}\). Evaluating these in the polynomial equation helps identify which, if any, of these rational numbers are indeed zeros of the polynomial.

Synthetic Division

Synthetic Division is a simplified form of polynomial division, especially useful to test whether a number is a root of a polynomial. It works well once you suspect a potential root, based on the Rational Root Theorem.
Consider the polynomial \( P(x) = 2x^3 + 7x^2 + 12x + 9 \). Suppose we test \( x = -1 \). In synthetic division, you start by writing down the coefficients: 2, 7, 12, and 9. Then, use the test root \( x = -1 \) in a calculation process that checks divisibility.

• Bring down the first coefficient (2) unchanged.
• Multiply this result by the test root (-1) and add to the next coefficient (7): \( 2 \times -1 + 7 = 5 \).
• Repeat for remaining coefficients: \( 5 \times -1 + 12 = 7 \), \( 7 \times -1 + 9 = 0 \).

A zero in the final spot confirms \( x = -1 \) is a root of the polynomial. The remaining part, \( 2x^2 + 5x + 9 \), is the quotient polynomial.

Complex Numbers

Complex numbers arise when solving polynomial equations with negative discriminants. They are numbers of the form \( a + bi \), where \( i \) is the imaginary unit and equals \( \sqrt{-1} \). In our polynomial example, after verifying \( x = -1 \) as a root, the remainder is the quadratic \( 2x^2 + 5x + 9 \).
To find its roots, we calculate the discriminant (\( b^2 - 4ac \)) of \( 2x^2 + 5x + 9 \), which results in -47, a negative number. This leads to complex roots.

• These complex roots are calculated as \( x = \frac{-5 \pm \sqrt{-47}}{4} \).
• This simplifies to \( x = \frac{-5 \pm i\sqrt{47}}{4} \), where \( i \) denotes the imaginary unit.

Thus, the polynomial equation yields complex conjugate roots.

Quadratic Formula

The Quadratic Formula provides a solution for any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For the quadratic \( 2x^2 + 5x + 9 \), we use this formula to determine the roots. Here, \( a = 2 \), \( b = 5 \), and \( c = 9 \).

• First, calculate the discriminant: \( b^2 - 4ac = 25 - 72 = -47 \).
• Since the discriminant is negative, the roots are complex numbers.
• Substituting, we get the roots: \( x = \frac{-5 \pm i\sqrt{47}}{4} \).

The formula effectively provides the solutions, indicating that no real zeros exist beyond the already identified rational zero. Thus, this method confirms the presence of complex solutions.