Problem 38 - College Algebra

Question

For the following exercises, find the zeros and give the multiplicity of each. $$ f(x)=3 x^{4}+6 x^{3}+3 x^{2} $$

Step-by-Step Solution

Answer

Zeros are \(x = 0\) with multiplicity 2 and \(x = -1\) with multiplicity 2.

Step 1: Factor the Polynomial

The first step is to factor out the greatest common factor (GCF) from the expression \( f(x) = 3x^4 + 6x^3 + 3x^2 \). The GCF in this case is \( 3x^2 \).Factoring \( 3x^2 \) out of the polynomial gives us:\[ f(x) = 3x^2(x^2 + 2x + 1) \]

Step 2: Factor the Quadratic

Next, focus on factoring the quadratic expression \( x^2 + 2x + 1 \). This can be rewritten as a perfect square trinomial:\[ x^2 + 2x + 1 = (x + 1)^2 \]Thus, the factored form of the polynomial becomes:\[ f(x) = 3x^2(x + 1)^2 \]

Step 3: Identify the Zeros

The zeros of the polynomial occur where the expression is equal to zero. From the factored form \( 3x^2(x + 1)^2 = 0 \), set each factor equal to zero:1. \( 3x^2 = 0 \) gives \( x = 0 \)2. \( (x + 1)^2 = 0 \) gives \( x = -1 \)

Step 4: Determine the Multiplicity

Multiplicity refers to the number of times a particular zero appears, based on the exponent in the factored form:1. The zero \( x = 0 \) comes from the factor \( 3x^2 \), so the multiplicity is 2.2. The zero \( x = -1 \) comes from the factor \( (x + 1)^2 \), so the multiplicity is also 2.

Key Concepts

Factoring Polynomials

Factoring polynomials is a key step in determining the roots or zeros of a polynomial function. It involves breaking down a complex polynomial into simpler factors that can be solved easily. Think of it as unpeeling an onion layer by layer, but instead of layers, you have factors that we need to identify and pull out.

In our case with the polynomial function \( f(x) = 3x^4 + 6x^3 + 3x^2 \), the first step is to find the greatest common factor (GCF). The GCF is the largest factor that divides all terms in the polynomial. Here, that factor is \( 3x^2 \), and factoring it out simplifies the polynomial to \( f(x) = 3x^2(x^2 + 2x + 1) \).

Once the GCF is factored out, we are left with a simpler quadratic expression \( x^2 + 2x + 1 \). This expression can often be factored further, leading us to discover even simpler expressions, sometimes perfect square trinomials, as we will see in the following sections.

Multiplicity of Roots

The multiplicity of a root gives us insight into the structure of a polynomial. It indicates how many times a particular solution appears, expressed as the exponent of its corresponding factor in the factored polynomial.

In the polynomial \( f(x) = 3x^2(x + 1)^2 \), after factoring, we find two roots: \( x = 0 \) and \( x = -1 \). Each of these roots is repeated or "counts" a certain number of times, which we describe using multiplicity.

• The zero at \( x = 0 \) comes from the factor \( 3x^2 \). Since the factor \( x \) is squared, it has a multiplicity of 2.
• Similarly, the zero at \( x = -1 \) derives from the factor \( (x + 1)^2 \). The squared factor tells us that this root also has a multiplicity of 2.

Understanding multiplicity helps predict the behavior of the polynomial function at each root, such as whether the graph touches or crosses the x-axis at those points.

Perfect Square Trinomial

A perfect square trinomial is a special type of quadratic expression that can be rewritten as the square of a binomial. Recognizing these trinomials makes factoring them much simpler, as you can quickly express them in their square form.

The quadratic part of our original problem, \( x^2 + 2x + 1 \), is an excellent example of a perfect square trinomial. It can be rewritten as \((x + 1)^2\) because:

• The first term \(x^2\) is a square of \(x\).
• The last term \(1\) is a square of \(1\).
• The middle term \(2x\) fits as the double product of the terms being squared, \(2(x)(1)\).

By identifying it as a perfect square trinomial, you can factor \(x^2 + 2x + 1\) swiftly into \((x + 1)^2\), simplifying the task of finding roots and further analysis of the polynomial. Recognizing and working with perfect squares makes many algebraic tasks easier and faster to complete, especially when dealing with polynomial roots.