Problem 5 - Algebra and Trigonometry

Question

\(1-12\) . A polynomial \(P\) is given. (a) Find all zeros of \(P\) , real and complex. (b) Factor \(P\) completely. $$ P(x)=x^{4}+2 x^{2}+1 $$

Step-by-Step Solution

Answer

Zeros: \( i, -i \). Factored: \((x^2 + 1)^2\).

Step 1: Recognize the Polynomial Structure

The polynomial \( P(x) = x^4 + 2x^2 + 1 \) can be seen as a quadratic in form if we substitute \( y = x^2 \). Then the polynomial becomes \( y^2 + 2y + 1 \).

Step 2: Factor the Quadratic

The quadratic \( y^2 + 2y + 1 \) can be factored as \( (y+1)(y+1) \), or \( (y+1)^2 \).

Step 3: Back-substitute to Original Variables

Since \( y = x^2 \), we substitute back to obtain \( (x^2 + 1)^2 \).

Step 4: Find the Zeros of the Factor

Set \( x^2 + 1 = 0 \). Solving gives \( x^2 = -1 \), hence \( x = i \) or \( x = -i \), where \( i \) is the imaginary unit, \( i = \sqrt{-1} \). These are the complex zeros of the polynomial.

Key Concepts

Factoring Polynomials

Factoring polynomials can be like solving a mystery. It involves breaking down an algebraic expression into simpler factors that multiply to make the original polynomial. In the polynomial from the original exercise, we notice that it is in the form of a quadratic, but with a twist—it involves a higher degree of 4.

By substituting variables, we identify the polynomial structure and factor it. Using the substitution method, as shown in the solution, we transformed the fourth-degree polynomial into a more familiar quadratic form:

• Original: \( P(x) = x^4 + 2x^2 + 1 \)
• Substitution: Let \( y = x^2 \).
• Quadratic form: \( y^2 + 2y + 1 \)

Breaking it into factors becomes more manageable:

• Factored form: \( (y+1)(y+1) \) or \( (y+1)^2 \).

Finally, by substituting back, we achieve the factors of the original polynomial in terms of \( x \) as \( (x^2 + 1)(x^2 + 1) \) or \( (x^2 + 1)^2 \). This lays the groundwork for finding zeros.

Complex Numbers

Complex numbers come into play when solving polynomial equations where the discriminant is negative, like when you face something like \( x^2 = -1 \). These numbers expand the number system to include solutions not covered by just real numbers.

To address this challenge, we leverage the imaginary unit \( i \), defined by \( i^2 = -1 \). This clever mathematical invention allows us to solve otherwise "unsolvable" equations.

In our exercise, after simplifying the polynomial to \( (x^2 + 1)^2 \), we set \( x^2 + 1 = 0 \). This implies:

• \( x^2 = -1 \)
• Thus, \( x = i \) (since \( i = \sqrt{-1} \)) or \( x = -i \)

These solutions are complex zeros of the original polynomial. Understanding these complex numbers helps in fully factoring polynomial expressions with all real and imaginary roots.

Quadratic Equations

Quadratic equations frequently appear in algebra and are often the gateway to higher mathematics. They typically take the form \( ax^2 + bx + c = 0 \). However, in our exercise, we saw a disguised quadratic in the form of a higher degree polynomial.

To solve these quadratics, factoring is a common method if the expression is "factorable". As seen, substitution makes our fourth-degree polynomial:

• Look like a quadratic: \( y^2 + 2y + 1 \) (where \( y = x^2 \)).

The quadratic equation \( y^2 + 2y + 1 = 0 \) factors neatly as \( (y+1)^2 = 0 \), revealing a double root \( y = -1 \). We then translate back to the original variable:

• \( x^2 = -1 \)
• Leading to complex solutions \( x = i \) and \( x = -i \)

Through these methods, not only do we find roots, but also we grasp the nature of polynomials’ relationships with their factors and roots.