Problem 29 - Algebra and Trigonometry

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 zeros are \(i\) and \(-i\), each with multiplicity 2.

Step 1: Check for Quadratic Form

The polynomial \(Q(x) = x^4 + 2x^2 + 1\) is in a quadratic form of \(x^2\): \((x^2)^2 + 2(x^2) + 1\). This is because the powers of \(x\) are double the power of \((x^2)\). Recognizing this will help in factoring the polynomial.

Step 2: Apply Quadratic Substitution

Let \(y = x^2\). Substitute \(y\) into the polynomial: \(y^2 + 2y + 1\). The aim is to factor this new quadratic expression.

Step 3: Factor the Quadratic Expression

Factor \(y^2 + 2y + 1\). This expression is a perfect square trinomial, which can be written as \((y + 1)^2\).

Step 4: Substitute Back to Original Variable

Revert the substitution by replacing \(y\) with \(x^2\): \((x^2 + 1)^2\). Now, the original polynomial \(Q(x) = (x^2 + 1)^2\).

Step 5: Solve for Zeros

Set the factored form equal to zero: \((x^2 + 1)^2 = 0\). Solving \(x^2 + 1 = 0\) gives us \(x^2 = -1\), which implies \(x = i\) or \(x = -i\).

Step 6: Determine Multiplicity of Zeros

Since \( (x^2 + 1)^2 \) is squared, each root \(x = i\) and \(x = -i\) has a multiplicity of 2.

Key Concepts

Quadratic Substitution

When working with polynomials, recognizing patterns is a key strategy to simplify complex expressions. Quadratic substitution is one such technique. By identifying a quadratic form within the expression, you can transform a high-degree polynomial into a simpler quadratic expression.
This is evident in our example: the polynomial \(x^4 + 2x^2 + 1\) looks daunting initially. However, by noticing that \(x^4\) can be written as \((x^2)^2\), the polynomial fits the pattern of a quadratic:\[(x^2)^2 + 2(x^2) + 1\].
To simplify, we introduce a substitution to convert it into a more recognizable quadratic form. Let \(y = x^2\). Substituting \(y\) into the polynomial, you get \(y^2 + 2y + 1\). Now it looks like a standard quadratic expression, making it much easier to factor.

Perfect Square Trinomial

A perfect square trinomial is a quadratic expression that can be factored into a binomial squared. In mathematical terms, if you have an expression of the form \(y^2 + 2y + 1\), it can be written as \((y + 1)^2\).
This recognition is crucial as it simplifies the factoring process. Spotting such patterns allows one to factor the expression quickly and accurately.

• Given the form \(a^2 + 2ab + b^2\), the trinomial \(y^2 + 2y + 1\) fits perfectly, where \(a = y\) and \(b = 1\).
• Thus, it simplifies to \((y + 1)^2\), showing it's a perfect square trinomial.

After factoring, you can substitute back to your original variable, making it easier to work with expressions in their simplest form.

Finding Zeros

Finding zeros of a polynomial is about determining the values of \(x\) for which the polynomial equals zero. For our example, once we substitute back \(y = x^2\), the polynomial \((x^2 + 1)^2\) is left to solve.
To find zeros, you set the equation equal to zero: \((x^2 + 1)^2 = 0\). We need to solve \(x^2 + 1 = 0\). This implies:

• \(x^2 = -1\)

Therefore, the solutions are complex numbers: \(x = i\) and \(x = -i\), where \(i\) is the imaginary unit. These are the zeros of our original polynomial, as they satisfy \(x^2 + 1 = 0\).

Multiplicity of Zeros

The multiplicity of a zero refers to the number of times a particular solution appears in the factored polynomial. When a factor is repeated, its zero's multiplicity increases.
In the expression \((x^2 + 1)^2\), each zero occurs due to a squared factor. Therefore, each root, both \(x = i\) and \(x = -i\), has a multiplicity of 2.

• This implies each zero appears twice in the polynomial's factorization.
• Understanding multiplicity helps in sketching the graph of the polynomial, as higher multiplicities affect the curve's shape at those points.

This wrapped our understanding of the polynomial: recognizing patterns, solving for zeros, and determining their multiplicities, enhancing our ability to handle polynomials.