Problem 13 - Precalculus Mathematics for Calculus

Question

A polynomial \(P\) is given. (a) Find all zeros of \(P\), real and complex. (b) Factor \(P\) completely. $$P(x)=x^{3}+8$$

Step-by-Step Solution

Answer

Zeros: \(-2, 1 + i\sqrt{3}, 1 - i\sqrt{3}\). Factor: \((x+2)(x^2-2x+4)\).

Step 1: Identify the Type of Polynomial

The given polynomial is a cubic polynomial with the form \( P(x) = x^3 + 8 \). It can be recognized as a sum of cubes.

Step 2: Apply the Sum of Cubes Formula

The sum of cubes formula is: \( a^3 + b^3 = (a+b)(a^2-ab+b^2) \). Here, \( a = x \) and \( b = 2 \), which allows us to rewrite \( x^3 + 8 \) as \( (x + 2)(x^2 - 2x + 4) \).

Step 3: Find the Real Zero

Set each factor to zero and solve: For \( x + 2 = 0 \), \( x = -2 \). This is the only real root of the polynomial.

Step 4: Find the Complex Zeros

Solve \( x^2 - 2x + 4 = 0 \) using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 1, b = -2, c = 4 \) into the formula to get \( x = \frac{2 \pm \sqrt{-12}}{2} \) which simplifies to \( x = 1 \pm i\sqrt{3} \). These are the complex roots.

Step 5: Write the Complete Factorization

The complete factorization of the polynomial is \( P(x) = (x + 2)(x - (1 + i\sqrt{3}))(x - (1 - i\sqrt{3})) \).

Key Concepts

Sum of Cubes Formula

The sum of cubes formula is a powerful tool for factoring cubic polynomials in the form of a sum of two cubes, such as \( x^3 + 8 \). This formula states that \( a^3 + b^3 = (a+b)(a^2-ab+b^2) \). In our example:

• \( a = x \)
• \( b = 2 \)

Substituting these values into the formula, we rewrite \( x^3 + 8 \) as \( (x + 2)(x^2 - 2x + 4) \). The formula simplifies the polynomial into factors that can be easily analyzed for real and complex roots.

Real Roots

Finding the real roots of a cubic polynomial like \( x^3 + 8 \) involves solving each factor of the expression obtained from the sum of cubes formula. The first factor, \( x + 2 \), is straightforward. Set it to zero to find the real root:

• \( x + 2 = 0 \)
• \( x = -2 \)

This indicates that \( x = -2 \) is the only real root of \( x^3 + 8 \). Real roots correspond to the points where the graph of the polynomial crosses the x-axis.

Complex Roots

After finding the real root, the next step is to identify complex roots by solving the remaining quadratic factor, \( x^2 - 2x + 4 \). We use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = 4 \).

• Calculate \( b^2 - 4ac = 4 - 16 = -12 \)
• Compute \( x = \frac{2 \pm \sqrt{-12}}{2} \)
• Simplify to \( x = 1 \pm i\sqrt{3} \)

The two complex roots are \( x = 1 + i\sqrt{3} \) and \( x = 1 - i\sqrt{3} \). These do not intersect the x-axis, representing instead the polynomial's behavior in the complex plane.

Polynomial Factorization

Completing the factorization of a cubic polynomial involves expressing it entirely in terms of its roots. For \( x^3 + 8 \), we already determined that:

• Real root: \( x = -2 \)
• Complex roots: \( x = 1 + i\sqrt{3} \) and \( x = 1 - i\sqrt{3} \)

Thus, the complete factorization is \( P(x) = (x + 2)(x - (1 + i\sqrt{3}))(x - (1 - i\sqrt{3})) \). This expression clearly shows all roots of the polynomial. Factorization helps in understanding the full landscape of solutions that satisfy the polynomial equation.