Problem 28 - Precalculus Mathematics for Calculus

Question

Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. $$P(x)=x^{6}-729$$

Step-by-Step Solution

Answer

Factor: \((x-9)(x+9)(x^2+9x+81)(x^2-9x+81)\); Zeros: 9 and -9, each with multiplicity 1.

Step 1: Recognize a Difference of Powers

The polynomial given is a difference of sixth powers: \[ P(x) = x^6 - 729 \]Recognize that 729 is a perfect sixth power, as \( 9^2 = 81 \) and \( 81^3 = 729 \). Therefore, this can be rewritten as:\[ P(x) = x^6 - 9^6 \]This is of the form \( a^6 - b^6 \).

Step 2: Apply the Difference of Squares Formula

The expression \( x^6 - 9^6 \) can be treated as a difference of squares:\[ (x^3)^2 - (9^3)^2 = (x^3 - 9^3)(x^3 + 9^3) \]This uses the identity: \( a^2 - b^2 = (a - b)(a + b) \).

Step 3: Factor Each Term Further Using Differences of Cubes

Now, factor each part using the formula for the differences of cubes:1. For \( x^3 - 9^3 \): \[ x^3 - 9^3 = (x - 9)(x^2 + 9x + 81) \]2. For \( x^3 + 9^3 \): \[ x^3 + 9^3 = (x + 9)(x^2 - 9x + 81) \]

Step 4: Verify Complete Factorization

Thus, the polynomial \( P(x) \) factors completely to:\[ (x - 9)(x^2 + 9x + 81)(x + 9)(x^2 - 9x + 81) \]Ensure each quadratic is irreducible over the reals by checking the discriminant for each one:1. \( x^2 + 9x + 81 \) has negative discriminant \( 9^2 - 4(1)(81) = 81 - 324 = -243 \)2. \( x^2 - 9x + 81 \) has negative discriminant \( (-9)^2 - 4(1)(81) = 81 - 324 = -243 \)Both quadratics are irreducible over the real numbers.

Step 5: Find Zeros and Their Multiplicities

Now, we find the zeros of \( P(x) \):1. \( (x - 9) = 0 \) gives \( x = 9 \).2. \( (x + 9) = 0 \) gives \( x = -9 \).Since the other factors are irreducible quadratics with no real roots, we only have real zeros:- Zero \( x = 9 \) with multiplicity 1.- Zero \( x = -9 \) with multiplicity 1.

Key Concepts

Difference of Powers

The concept of the difference of powers is a useful tool when factoring polynomials like the given exercise, where we have a sixth-degree polynomial: \( P(x) = x^6 - 729 \). In this case, we recognize that 729 is not only a power of 6 but is exactly \( 9^6 \). Thus, transforming the expression into \( x^6 - 9^6 \) is key. This allows us to use other factoring techniques.

When dealing with differences of powers, our main target is to break down the expression to simpler terms, using identities like the difference of squares, cubes, and more. Always ensure to simplify expressions at each step to fully exploit these useful identities.

Difference of Squares

The difference of squares formula is a powerful method for factoring expressions where two square terms are subtracted. This method was used in the solution when we approached the equation \( (x^3)^2 - (9^3)^2 \) from the polynomial \( x^6 - 9^6 \). The difference of squares identity is:

\[ a^2 - b^2 = (a - b)(a + b) \]

Applying this identity transformed the equation into \( (x^3 - 9^3)(x^3 + 9^3) \), breaking down the powers and making them easier to handle with further factorization techniques. This process reduces complex power differences into factors that may further simplify, or occasionally leave non-factorable quadratics or other forms.

Difference of Cubes

Next, we apply the difference of cubes formula, an adaptation in factoring that helps us deal with terms like \( x^3 - 9^3 \) from our factorization. The identities used are:

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

These principles were crucial in transforming \( x^3 - 9^3 \) into \((x - 9)(x^2 + 9x + 81)\) and \( x^3 + 9^3 \) into \((x + 9)(x^2 - 9x + 81)\). By applying these identities, we break down cubic differences further into irreducible polynomials over the reals unless the discriminant suggests otherwise.

Zeros of a Polynomial

Zeros of a polynomial are values that make the polynomial equal to zero. In the context of the factored polynomial \( (x - 9)(x^2 + 9x + 81)(x + 9)(x^2 - 9x + 81) \), identifying zeros means solving each linear factor.

- For \( (x - 9) = 0 \), we deduce a zero at \( x = 9 \).
- For \( (x + 9) = 0 \), the zero is \( x = -9 \).

The quadratic factors yield complex roots, which do not contribute any more real zeros. As a result, the polynomial \( P(x) \) has real zeros at \( x = 9 \) and \( x = -9 \), both appearing once in the factorization.

Multiplicity of Zeros

Multiplicity refers to how many times a specific zero appears in the polynomial. In this problem, we evaluated multiplicities by referring to the factored form.

The linear factors found in \((x - 9)(x + 9)\) indicate each zero has a multiplicity of 1. Multiplicity plays an essential role in polynomial behavior at zeros:

• A zero with odd multiplicity crosses the x-axis.
• A zero with even multiplicity only touches the x-axis without crossing.

The knowledge of a zero's multiplicity aids in graphing and understanding the polynomial's nature.