Problem 9 - Algebra and Trigonometry

Question

\(1-12\) . 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 of \(P(x)\) are \(-2, 1+i\sqrt{3}, 1-i\sqrt{3}\); factorization is \((x+2)(x-(1+i\sqrt{3}))(x-(1-i\sqrt{3}))\).

Step 1: Rewriting as a difference of cubes

The polynomial can be rewritten as a difference of cubes. Recognize that \(x^3 + 8\) can be expressed as \(x^3 + 2^3\). This matches the form \(a^3 + b^3\) where \(a = x\) and \(b = 2\).

Step 2: Applying the sum of cubes formula

Use the sum of cubes identity: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). Substituting \(a = x\) and \(b = 2\), we get: \(P(x) = (x + 2)(x^2 - 2x + 4)\).

Step 3: Finding real zeros

The expression \((x + 2)\) gives us the real zero of the polynomial. Therefore, the real zero is \(x = -2\).

Step 4: Solving for complex zeros

To find the complex zeros, solve the quadratic equation \(x^2 - 2x + 4 = 0\) using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = -2\), and \(c = 4\).

Step 5: Calculating the discriminant

Compute the discriminant: \(b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot 4 = 4 - 16 = -12\). Since the discriminant is negative, the quadratic equation has two complex solutions.

Step 6: Finding the complex solutions

Using the quadratic formula with the negative discriminant, calculate: \(x = \frac{2 \pm \sqrt{-12}}{2}\). Simplify \(\sqrt{-12}\) as \(i\sqrt{12} = 2i\sqrt{3}\). Thus, the roots are \(x = 1 \pm i\sqrt{3}\).

Step 7: Stating the zeros and factoring

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

Key Concepts

Difference of Cubes

A cube in mathematics is the result of multiplying a number or variable by itself twice. When we have a polynomial that fits a three-term structure of the form \(x^3 + a^3\) or \(x^3 - a^3\), it's either a **sum of cubes** or **difference of cubes**. In this exercise, we have the polynomial \(x^3 + 8\), which is a perfect example of a sum of cubes, as it can be rewritten as \(x^3 + 2^3\). Recognizing this form allows us to apply specific algebraic formulas to factor and solve it effectively. The essential part when identifying a difference or sum of cubes is spotting if both terms are perfect cubes themselves.

Sum of Cubes Formula

The **sum of cubes formula** is a useful tool in algebra that allows us to factor expressions like \(a^3 + b^3\). The formula states:

• \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

This identity breaks the cubic expression into simpler polynomial factors that are easier to handle. For our polynomial \(P(x) = x^3 + 8\), we let \(a = x\) and \(b = 2\). Applying the formula, we get:

• \((x + 2)(x^2 - 2x + 4)\)

Now, the polynomial is separated into a linear part \((x + 2)\), and a quadratic part \((x^2 - 2x + 4)\), further simplifying the problem of finding zeros.

Real Zeros

Real zeros of a polynomial are the values of \(x\) that make the polynomial equal to zero. These are the x-intercepts on a graph where the curve crosses the x-axis. When we factor \(P(x) = x^3+8\) as \((x + 2)(x^2 - 2x + 4)\), the real zero comes directly from the linear factor:

• \(x + 2 = 0 \Rightarrow x = -2\)

This tells us that one real zero of the polynomial is \(x = -2\). At this point, we've discovered all the points where the polynomial crosses the x-axis. The other zeros will be from the quadratic factor, which might involve complex numbers.

Complex Zeros

Complex zeros are solutions to polynomial equations that aren't visible on the real number line. They often arise from factors that don't produce real solutions, like the quadratic we see in our exercise, \(x^2 - 2x + 4\). To find these complex zeros, we apply the **quadratic formula**:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

With \(a = 1\), \(b = -2\), and \(c = 4\), we calculate the discriminant \(b^2 - 4ac = -12\). Since it's negative, the roots are complex. They are calculated as:

• \(x = 1 \pm i\sqrt{3}\)

Thus, these are the complex zeros of the polynomial. This process reveals the hidden complexity that isn't visible through simple graphing.