Problem 11 - Algebra and Trigonometry

Question

\(1-12\) . A polynomial \(P\) is given. (a) Find all zeros of \(P\) , real and complex. (b) Factor \(P\) completely. $$ P(x)=x^{6}-1 $$

Step-by-Step Solution

Answer

Zeros: \(x = 1, -1, \frac{-1 \pm i \sqrt{3}}{2}, \frac{1 \pm i \sqrt{3}}{2}\); Factor: \((x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1)\).

Step 1: Recognize the Expression

The given polynomial is \( P(x) = x^6 - 1 \), which resembles a difference of two squares. This is a known identity that can be written as \( a^2 - b^2 = (a-b)(a+b) \).

Step 2: Decompose as a Difference of Squares

Given that \( x^6 - 1 = (x^3)^2 - 1^2 \), apply the difference of squares formula: \((x^3 - 1)(x^3 + 1)\).

Step 3: Further Factor Each Expression

Factor \( x^3 - 1 \) using the difference of cubes formula: \((x - 1)(x^2 + x + 1)\). Factor \( x^3 + 1 \) using the sum of cubes formula: \((x + 1)(x^2 - x + 1)\).

Step 4: Collect All Factors

Combine all factors: \( P(x) = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1) \). This expression shows the polynomial completely factored.

Step 5: Find Real Zeros

Set each linear factor to zero: \(x - 1 = 0\) leads to \(x = 1\) and \(x + 1 = 0\) leads to \(x = -1\). So we have two real zeros.

Step 6: Solve Quadratic Equations for Complex Zeros

For \(x^2 + x + 1 = 0\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1, b = 1, c = 1\). Discriminant is \(b^2 - 4ac = 1 - 4 = -3\), leading to complex zeros \(x = \frac{-1 \pm i \sqrt{3}}{2}\). For \(x^2 - x + 1 = 0\), apply the quadratic formula similarly to find \(x = \frac{1 \pm i \sqrt{3}}{2}\).

Step 7: List All Zeros

The zeros of \(P(x)\) are \(x = 1, x = -1, x = \frac{-1 \pm i \sqrt{3}}{2}\), and \(x = \frac{1 \pm i \sqrt{3}}{2}\).

Key Concepts

Difference of Squares

In algebra, a difference of squares is a fundamental concept that helps in factoring polynomials. It specifically involves expressions of the form \( a^2 - b^2 \). This pattern can be rewritten using the identity \( a^2 - b^2 = (a-b)(a+b) \).

When you see something like \( x^6 - 1 \), which was part of our given problem, you can imagine it as \( (x^3)^2 - 1^2 \). This transforms the problem into the difference of two squares.

• First, recognize the expression could be something squared minus another square.
• Then apply the difference of squares formula to break it down.

Using our example, we have \( (x^3)^2 - 1^2 = (x^3 - 1)(x^3 + 1) \), which simplifies our expression into two more manageable parts. Next steps will involve dealing with these new expressions.

Difference of Cubes

The difference of cubes is another critical technique in polynomial factoring. This applies to expressions like \( a^3 - b^3 \), where it can be factored using the formula \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \).

In our exercise, one part became \( x^3 - 1 \), which fits the difference of cubes pattern:

• Identify it as a cube minus another cube: \( x^3 - 1^3 \).
• Split it using the formula to get \( (x - 1)(x^2 + x + 1) \).

Understanding this concept helps break down complicated expressions into linear and quadratic factors, making it easier to factor them completely.

Sum of Cubes

The sum of cubes is similar yet distinct from the difference of cubes. It applies to expressions like \( a^3 + b^3 \), and you use the formula \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \).

In our task, after factoring the difference of squares, one of the expressions became \( x^3 + 1 \). This fits the sum of cubes pattern:

• Recognize as a sum of cubes: \( x^3 + 1^3 \).
• Apply the formula: this gives us \( (x + 1)(x^2 - x + 1) \).

Utilizing the sum of cubes formula enables further simplification, which is crucial for full factorization of polynomials.

Complex Zeros

Finding zeros of a polynomial, especially complex zeros, requires solving equations where the roots may not be real numbers. In our problem, after reducing the polynomial, we encountered quadratic expressions that do not factor easily into real numbers only.

For these, we use the quadratic formula:

• For \( x^2 + x + 1 = 0 \), the formula gives complex solutions, since the discriminant \( 1 - 4 \) is negative.
• This yields zeros \( x = \frac{-1 \pm i\sqrt{3}}{2} \).
• Similarly, \( x^2 - x + 1 = 0 \) provides \( x = \frac{1 \pm i\sqrt{3}}{2} \).

Complex numbers generally appear as pairs and are essential in expressing the complete set of solutions for polynomials. Understanding this concept helps in identifying all the zeros, whether real or complex, of a polynomial expression.