Problem 28 - Algebra and Trigonometry

Question

Factor the polynomial and use the factored form to find the zeros. Then sketch the graph. $$ P(x)=x^{5}-9 x^{3} $$

Step-by-Step Solution

Answer

Zeros: \( x = -3, 0, 3 \). Factored form: \( x^3(x - 3)(x + 3) \).

Step 1: Identify the Greatest Common Factor (GCF)

The polynomial given is \( P(x) = x^5 - 9x^3 \). First, identify the greatest common factor in both terms. Both terms share the factor \( x^3 \), so the GCF is \( x^3 \).

Step 2: Factor Out the GCF

Factor \( x^3 \) out of the polynomial. This gives us:\[ x^3(x^2 - 9) \] which is an intermediate factored form of the polynomial.

Step 3: Factor the Quadratic Expression

Next, focus on the quadratic expression \( x^2 - 9 \). Recognize that this is a difference of squares. Factor it as follows:\[ x^2 - 9 = (x - 3)(x + 3) \].

Step 4: Write the Completely Factored Form

Substitute the factorization from the previous step back into the expression:\[ P(x) = x^3(x - 3)(x + 3) \]. This is the completely factored form of the polynomial.

Step 5: Find the Zeros from the Factored Form

To find the zeros of the polynomial, set each factor equal to zero:1. \( x^3 = 0 \) gives \( x = 0 \).2. \( x - 3 = 0 \) gives \( x = 3 \).3. \( x + 3 = 0 \) gives \( x = -3 \).Thus, the zeros are \( x = 0 \), \( x = 3 \), and \( x = -3 \).

Step 6: Sketch the Graph

The graph of \( P(x) = x^5 - 9x^3 \) crosses the x-axis at the points \( x = -3, 0, 3 \). Since the leading term is \( x^5 \), the end behavior indicates that as \( x \to -\infty \), \( P(x) \to -\infty \), and as \( x \to \infty \), \( P(x) \to \infty \). Sketch the graph starting from the bottom left, crossing the x-axis at \( x = -3 \), remaining in the third quadrant before touching or crossing at the origin (since \( x^3 \) multiplicity causes a bounce), and going up through \( x = 3 \).

Key Concepts

Greatest Common Factor (GCF)

The greatest common factor, or GCF, is the largest factor that divides each term of a polynomial. Finding the GCF is the first step in many factoring problems, especially when working with polynomials. In the polynomial \( P(x) = x^5 - 9x^3 \), the process begins by identifying common factors in each term. Both terms, \( x^5 \) and \(-9x^3 \), share the factor \( x^3 \).

Therefore, the GCF here is \( x^3 \). This plays a crucial role in simplifying the polynomial, making it easier to further factor or solve. By factoring out \( x^3 \) from each term, we express the polynomial in an easier form:

• The first term \( x^5 \) becomes \( x^2 \) after factoring out \( x^3 \).
• The second term \(-9x^3 \) becomes \(-9 \).

Thus, by factoring out \( x^3 \), we rewrite the polynomial as \( x^3(x^2 - 9) \). This transformation prepares it for further factoring and solving steps.

Zeros of a Polynomial

Finding the zeros of a polynomial involves setting the equation equal to zero and solving for the values of \( x \). To do this efficiently, having the polynomial in its factored form helps a great deal. In factored form, such as \( P(x) = x^3(x - 3)(x + 3) \) for our example, the problem simplifies:

Each factor equated to zero leads to the zeros of the polynomial. Here, we have three factors:

• \( x^3 = 0 \), solving this gives \( x = 0 \). The zero \( x = 0 \) is of multiplicity 3, meaning it affects the shape of the graph, typically causing a bounce or touch at the x-axis.
• \( x - 3 = 0 \), solving gives \( x = 3 \).
• \( x + 3 = 0 \), solving gives \( x = -3 \).

Thus, the polynomial \( P(x) \) has zeros at \( x = -3, 0, \) and \( 3 \). Each zero corresponds to an x-intercept on the graph of the polynomial.

Difference of Squares

The difference of squares is a special pattern observed in algebra where two terms squared are subtracted. It takes the form \( a^2 - b^2 \) and can be factored into \((a - b)(a + b)\). It is a handy technique for factoring certain polynomials.

In the expression \( x^2 - 9 \) from the polynomial \( P(x) \), we see this pattern:

• Here, \( a^2 \) is \( x^2 \), and \( b^2 \) is \( 9 \) (or \( 3^2 \)).

Recognizing this, we can factor \( x^2 - 9 \) into \((x - 3)(x + 3)\). This step reveals hidden linear factors, allowing us to completely factor the polynomial and find zeros more easily.

Handling expressions as a difference of squares significantly simplifies working on polynomial equations and plays a crucial role in unveiling their structure and solutions.