Problem 29 - Precalculus Mathematics for Calculus

Question

Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. $$Q(x)=x^{4}+2 x^{2}+1$$

Step-by-Step Solution

Answer

The polynomial factors as \((x^2+1)^2\). Zeros are \(i\) and \(-i\), each with multiplicity 2.

Step 1: Identify the polynomial form

The given polynomial is \( Q(x) = x^4 + 2x^2 + 1 \). Notice that this is a quadratic in terms of \( x^2 \), since letting \( y = x^2 \) transforms it into \( y^2 + 2y + 1 \).

Step 2: Factor the transformed quadratic

The quadratic \( y^2 + 2y + 1 \) is a perfect square trinomial. It can be factored as \( (y + 1)^2 \).

Step 3: Substitute back in terms of x

Substituting back \( y = x^2 \), the polynomial \( (y + 1)^2 \) becomes \( (x^2 + 1)^2 \). Hence, \( Q(x) = (x^2 + 1)^2 \).

Step 4: Find the zeros of the factor

To find the zeros of \( Q(x) = (x^2 + 1)^2 \), set \( x^2 + 1 = 0 \). Solving for \( x \), we get \( x^2 = -1 \), which implies \( x = i \) and \( x = -i \), where \( i \) is the imaginary unit.

Step 5: Determine the multiplicity of each zero

Since \( (x^2 + 1)^2 \) means the factor \( x^2 + 1 \) is repeated twice, the zeros \( i \) and \( -i \) each have a multiplicity of 2.

Key Concepts

Zeros of Polynomials

The zeros of a polynomial are the values of the variable that satisfy the equation when the polynomial is set to zero. In simple terms, they are the solutions to the equation formed by equating the polynomial expression to zero.
For example, if we have a polynomial \( Q(x) = x^4 + 2x^2 + 1 \), the zeros are the values of \( x \) for which \( Q(x) = 0 \). In the solution, we transformed it into \( (x^2 + 1)^2 \). Setting this equal to zero, \( x^2 + 1 = 0 \) leads us to solve \( x^2 = -1 \).
To find the zeros, we have to solve this equation. It's important to note that not all zeros are real numbers; they can sometimes be complex numbers, which we will address later on. For the given polynomial, the zeros are \( i \) and \( -i \), because these values of \( x \) satisfy \( x^2 = -1 \). Identifying zeros is a key part in understanding the behavior and graphical representation of polynomials.

Multiplicity of Zeros

Multiplicity refers to the number of times a particular zero appears as a solution to a polynomial equation. When looking at the factorization of the polynomial, multiplicity is indicated by the power to which a factor is raised.
Let’s consider the factorization \( Q(x) = (x^2 + 1)^2 \). The factor \( (x^2 + 1) \) is squared. This means each zero of the equation \( x^2 + 1 = 0 \) has a multiplicity of 2. In other words, the zeros \( i \) and \( -i \) each appear twice in the solution set.
Understanding the multiplicity of zeros is crucial when it comes to graphing polynomials because it impacts the behavior of the graph at the zero. A zero with an even multiplicity will cause the graph to "bounce" off the x-axis at that point, while an odd multiplicity will cause the graph to "pass through" the axis.

Complex Numbers

Complex numbers arise naturally in mathematics, especially when dealing with polynomials and their zeros. They are numbers that have a real part and an imaginary part and are usually expressed in the form \( a + bi \), where \( a \) and \( b \) are real numbers and \( i \) is the imaginary unit defined as \( i = \sqrt{-1} \).
In the context of the polynomial \( Q(x) = (x^2 + 1)^2 \), the equation \( x^2 = -1 \) doesn't have real solutions because no real number, when squared, results in \(-1\). However, using complex numbers, we find that \( x = i \) and \( x = -i \) satisfy this equation.
Complex numbers allow us to find solutions to polynomial equations that have no solutions in the real numbers alone. They are an expansion of the number system that enables us to solve equations like \( x^2 + 1 = 0 \), which cannot be handled by real numbers alone. Understanding complex numbers widens our ability to work with a range of mathematical problems.