Problem 17 - Precalculus Mathematics for Calculus

Question

Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. $$P(x)=x^{2}+25$$

Step-by-Step Solution

Answer

Zeros are \( x = 5i \) and \( x = -5i \), both with multiplicity 1.

Step 1: Recognize the Form

The polynomial \( P(x) = x^2 + 25 \) is a sum of squares. Recall that it can be rewritten as \( x^2 + (5i)^2 \). There is no real number whose square is negative, but we can factor this using complex numbers.

Step 2: Use the Sum of Squares Formula

For the sum of squares, \( a^2 + b^2 \), we use the formula \((a + bi)(a - bi)\). Here, \( a = x \) and \( b = 5i \). So, we factor \( P(x) \) as \( (x + 5i)(x - 5i) \).

Step 3: Identify Zeros from Factors

Set each factor equal to zero to find the zeros.\( \begin{align*} x + 5i &= 0 \rightarrow x = -5i, \ x - 5i &= 0 \rightarrow x = 5i.\end{align*} \)

Step 4: Determine Multiplicities

Each zero \( x = -5i \) and \( x = 5i \) comes from a distinct linear factor, so both have a multiplicity of 1.

Key Concepts

Polynomial Factoring

Polynomial factoring is a crucial concept in algebra, which involves expressing a polynomial as a product of simpler polynomials. This simplification helps in solving equations and understanding polynomial properties. In the polynomial given, \( P(x) = x^2 + 25 \), identifying it as a sum of squares helps recognize that traditional factoring methods need to incorporate complex numbers. When we rewrite \( P(x) \) as \( x^2 + (5i)^2 \), we use complex numbers because the real number system does not allow the square root of a negative number.

• First, note that \( x^2 + (5i)^2 \) is a sum of squares, where \( a = x \) and \( b = 5i \).
• The sum of squares can be factored using the formula \((a + bi)(a - bi)\), yielding \( (x + 5i)(x - 5i) \).

This approach provides insight into how polynomials containing imaginary components behave and are factored, expanding our understanding beyond only real numbers.

Sum of Squares

The sum of squares is an important concept in algebra, especially when dealing with polynomials that do not fit typical factoring scenarios. The sum \( a^2 + b^2 \) usually cannot be factored over the reals, but with complex numbers, it is possible. In the context of the polynomial \( P(x) = x^2 + 25 \), we recognize it as \( x^2 + (5i)^2 \), emphasizing its applicability to complex numbers.

• The factorization uses the sum of squares formula: \((a + bi)(a - bi)\).
• In our example, substituting \( a = x \) and \( b = 5i \) gives the product \( (x + 5i)(x - 5i) \).

This converts the non-factorable "sum" of squares into factors easily used for determining zeros. It reveals how complex conjugates work together to explain polynomial roots, even when these roots aren't real.

Zeros and Multiplicity

Zeros of a polynomial are the values of \( x \) that make the polynomial equal to zero. Identifying these zeros, especially with their multiplicities, helps us understand more about the function's graph and behavior. In our exercise, the zeros arise from setting each factor of \( (x + 5i)(x - 5i) \) to zero.

• From \( x + 5i = 0 \), the zero is \( x = -5i \).
• From \( x - 5i = 0 \), the zero is \( x = 5i \).

Each zero occurs once, meaning they have a multiplicity of 1. This means that each root is simple, not repeated. The multiplicity informs us that the polynomial does not touch or bounce off the imaginary axis; it crosses it cleanly at the given zeros, illustrating clear separations.