Problem 52 - Precalculus Mathematics for Calculus

Question

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

Step 1: Check for Obvious Zeros

First, we test if there are any obvious zeros by substituting simple values for \( x \), such as \( x = 0, 1, -1 \). For this polynomial, substituting these values doesn't yield a zero. Move on to the Rational Root Theorem.

Step 2: Use the Rational Root Theorem

The Rational Root Theorem suggests that any rational root, in its simplest form, is a fraction \( \frac{p}{q} \) where \( p \) is a factor of the constant term (-9) and \( q \) is a factor of the leading coefficient (2). Possible rational roots are thus \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2} \).

Step 3: Test Possible Rational Roots

Test these possible roots in the polynomial. By substitution, we find that \( x = 3 \) is a root because \( P(3) = 0 \).

Step 4: Polynomial Division

Since \( x = 3 \) is a root, divide \( P(x) \) by \( x - 3 \) using synthetic or long division to simplify the polynomial. The division results are \( 2x^2 - 2x + 3 \).

Step 5: Solve the Quadratic Equation

Set the simplified polynomial equal to zero: \( 2x^2 - 2x + 3 = 0 \). Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = -2, c = 3 \).

Step 6: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac = (-2)^2 - 4(2)(3) = 4 - 24 = -20 \). Since the discriminant is negative, there are no real roots.

Step 7: Find Complex Solutions

With a negative discriminant, the solutions are complex. Using the quadratic formula: \( x = \frac{2 \pm \sqrt{-20}}{4} = \frac{2 \pm 2i\sqrt{5}}{4} = \frac{1 \pm i\sqrt{5}}{2} \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a useful tool for finding the zeros of a polynomial. It gives us a systematic way to list possible rational roots of a polynomial equation.
Here's how it works:

• First, identify the constant term and the leading coefficient of your polynomial. In the example polynomial, \( P(x) = 2x^3 - 8x^2 + 9x - 9 \), the constant term is \(-9\) and the leading coefficient is \(2\).
• The possible rational roots are fractions \( \frac{p}{q} \), where \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient.
• For \(-9\), the factors are \(\pm 1, \pm 3, \pm 9\), and for \(2\), they are \(\pm 1, \pm 2\).
• Thus, the potential rational roots are \(\pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}\).

This method narrows down potential candidates for the polynomial's zeros, allowing us to check each one until we find an actual root that makes the polynomial equal to zero.

Polynomial Division

Once a possible zero of a polynomial is identified, polynomial division becomes the next logical step to simplify the polynomial. Suppose we find \( x = 3 \) as a root for our polynomial \( P(x) \). Polynomial division helps us divide \( P(x) \) by \( x - 3 \).There are two common methods:

• Long Division: Similar to numerical long division, align terms by decreasing powers and divide step by step.
• Synthetic Division: Requires less writing space and handles the coefficients directly, offering a quicker computation for polynomials.

Through either method, dividing \( P(x) = 2x^3 - 8x^2 + 9x - 9 \) by \( x - 3 \) results in the quotient \( 2x^2 - 2x + 3 \). This new polynomial is of a lower degree, making it easier to solve and continue the search for the remaining zeros.

Complex Solutions

When solving for the zeros of a polynomial, finding a negative discriminant indicates the presence of complex solutions. In the example, after using polynomial division, we have the quadratic \( 2x^2 - 2x + 3 = 0 \).To find the zeros, apply the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \( a = 2 \), \( b = -2 \), and \( c = 3 \). Calculate the discriminant:\[b^2 - 4ac = (-2)^2 - 4(2)(3) = 4 - 24 = -20\]With a negative discriminant, the roots are complex. Substitute back into the quadratic formula to solve:\[ x = \frac{2 \pm \sqrt{-20}}{4} = \frac{2 \pm 2i\sqrt{5}}{4} = \frac{1 \pm i\sqrt{5}}{2}\]Thus, the polynomial has two complex zeros: \( \frac{1 + i\sqrt{5}}{2} \) and \( \frac{1 - i\sqrt{5}}{2} \). These complex roots show up as a conjugate pair due to the nature of the quadratic equation with real coefficients.