Problem 17 - Precalculus: Functions and Graphs

Question

Find the zeros of \(f(x),\) and state the multiplicity of each zero. $$f(x)=4 x^{5}+12 x^{4}+9 x^{3}$$

Step-by-Step Solution

Answer

The zeros are \(x = 0\) (multiplicity 3) and \(x = -\frac{3}{2}\) (multiplicity 2).

Step 1: Factor out the Greatest Common Factor (GCF)

First, factor out the greatest common factor from all terms in the polynomial. The greatest common factor of the terms in \(f(x) = 4x^5 + 12x^4 + 9x^3\) is \(x^3\). This gives us \(f(x) = x^3(4x^2 + 12x + 9)\).

Step 2: Factor the Quadratic Expression

Next, factor the quadratic expression \(4x^2 + 12x + 9\) in the parentheses. We look for two numbers that multiply to \(4 imes 9 = 36\) and add up to \(12\). These numbers are \(6\) and \(6\). This allows us to rewrite the quadratic as \((2x + 3)^2\), giving us \(f(x) = x^3(2x + 3)^2\).

Step 3: Identify the Zeros of the Expression

The zero of a polynomial is found by setting the factored expression equal to zero. Set \(x^3(2x + 3)^2 = 0\). The zeros occur where each factor equals zero. Therefore, solve \(x^3 = 0\) and \((2x + 3)^2 = 0\).

Step 4: Solve for Each Zero

For \(x^3 = 0\), the solution is \(x = 0\) (since the cube of zero is zero). For \((2x + 3)^2 = 0\), solve \(2x + 3 = 0\) for \(x\): \(2x = -3\) so \(x = -\frac{3}{2}\).

Step 5: Determine the Multiplicity of Each Zero

The multiplicity of a zero corresponds to its power in the factorization. For \(x = 0\), since the factor is \(x^3\), the multiplicity is 3. For \(x = -\frac{3}{2}\), since the factor is \((2x + 3)^2\), the multiplicity is 2.

Key Concepts

Greatest Common Factor

A crucial first step in dealing with polynomials is finding the Greatest Common Factor (GCF). The GCF of a polynomial is the largest factor that divides each term of the polynomial without leaving a remainder. In the case of the polynomial \( f(x) = 4x^5 + 12x^4 + 9x^3 \), the GCF helps simplify the expression, making it easier to solve for the zeros.

To find the GCF:

• List the factors of each term.
• Identify the common factors.
• Select the largest factor present in each term.

In our polynomial, the GCF is \(x^3\). This is because each term (\(4x^5\), \(12x^4\), and \(9x^3\)) has at least \(x^3\) as a factor. By factoring out \(x^3\), we reduce the polynomial to \( f(x) = x^3 (4x^2 + 12x + 9) \). This simplification makes subsequent steps more manageable.

Quadratic Factoring

Once the common factor has been removed, the next task is often to factor the remaining expression, particularly if it's a quadratic. Quadratic factors take the form \(ax^2 + bx + c\). The goal is to rewrite it as a product of two binomials.

For the expression \(4x^2 + 12x + 9\), we'll look for two numbers that multiply to \(4 \times 9 = 36 \) and add up to the middle coefficient, 12. These numbers are 6 and 6 (since \(6 \times 6 = 36\) and \(6 + 6 = 12\)).

This allows us to express the quadratic as \((2x + 3)^2\), making the complete factorization \( f(x) = x^3(2x + 3)^2 \). Quadratic factoring simplifies solving for zeros because we can see the expression in terms of previously learned basic equations.

Multiplicity of Zeros

The multiplicity of a zero relates to how many times a particular factor appears in the polynomial's factorization. It tells us whether the graph of the polynomial touches or crosses the x-axis at these points.

For our polynomial, \( f(x) = x^3 (2x + 3)^2 \), zero values arise from setting each factor to zero:

• The factor \(x^3\) gives a zero of \(x = 0\) with multiplicity 3.
• The factor \((2x + 3)^2\) results in the zero \(x = -\frac{3}{2}\) with multiplicity 2.

The multiplicity has geometrical interpretations:

• A zero with an odd multiplicity (like \(x=0\), multiplicity 3) indicates the curve crosses the x-axis.
• A zero with an even multiplicity (like \(x=-\frac{3}{2}\), multiplicity 2) means the curve only touches the x-axis at this point and then turns back.

Understanding multiplicity gives insight into the shape and behavior of the polynomial's graph.