Problem 41 - College Algebra

Question

In Exercises \(37-44,\) find all the zeros of the function and write the polynomial as a product of linear factors. $$ f(x)=x^{4}+37 x^{2}+36 $$

Step-by-Step Solution

Answer

The zeros of the given polynomial are \(±6i\) and \(±i\). The polynomial in terms of linear factors is \([(x-6i)(x+6i) (x-i)(x+i)]\).

Step 1: Solve the polynomial equation

We set the polynomial equal to zero and find its solutions, these are the zeros of the polynomial.\n\[x^{4}+37x^{2}+36=0\]

Step 2: Factor the equation

Look for two numbers that add up to 37 and multiply to 36. These numbers are 36 and 1. Now the equation can be factored as follows:\n\[(x^{2}+36)(x^{2}+1)=0\]

Step 3: Solve the factored equation

Now, we set each factor equal to zero and solve for x:\n\[x^{2}+36 = 0 => x^{2} = -36 => x=±6i\]\n\[x^{2}+1 = 0 => x^{2} = -1 => x=±i\]

Step 4: Rewrite the polynomial as a product of linear factors

Now that we have all the zeros of f(x), the polynomial can be written as a product of linear factors.\n\[f(x) = [(x-6i)(x+6i)][(x-i)(x+i)]\]

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial skill in algebra, which allows us to break down polynomial equations into simpler, more manageable parts. When we factor a polynomial, we express it as a product of its polynomial factors. This process simplifies solving equations and identifying the polynomial's zeros.
One common method to factor polynomials is by finding two numbers that satisfy specific conditions relating to the coefficients of the polynomial. For example, when factoring the polynomial \(x^4 + 37x^2 + 36\), we looked for numbers that add up to 37 and multiply to 36. These numbers were identified as 36 and 1.
Once these numbers are found, they are used to split the middle term and allow for grouping, eventually factoring the polynomial into simpler binomial factors. For the given problem, the polynomial was successfully factored into \((x^2 + 36)(x^2 + 1)\). This transformation is vital as it sets the stage for solving the polynomial equation to find its zeros.

Imaginary Numbers

Imaginary numbers arise when we encounter the square root of a negative number. They are represented using the symbol \(i\), where \(i^2 = -1\). Encountering imaginary numbers in a polynomial usually indicates that the polynomial's graph will not intersect the x-axis at these points, as they are not real zeros.
In the given exercise, we encounter imaginary numbers when solving the factors \(x^2 + 36 = 0\) and \(x^2 + 1 = 0\). Solving these gives us roots of \(x = \pm 6i\) and \(x = \pm i\), respectively.
Understanding imaginary numbers is essential for solving polynomials, as it expands the set of possible solutions, providing a complete picture of where and how a polynomial can equal zero. Without imaginary numbers, solutions to some polynomial equations would remain unknown, limiting mathematical exploration and understanding.

Solving Polynomial Equations

To solve polynomial equations, we typically set the polynomial equal to zero and then identify solutions, also called zeros or roots. These zeros are values of \(x\) that make the polynomial evaluate to zero.
After factoring the polynomial \(f(x)=x^4+37x^2+36\) to \((x^2+36)(x^2+1)=0\), we solve each factor separately. Setting each factor equal to zero, \(x^2+36=0\) and \(x^2+1=0\), leads to the solutions \(x = \pm 6i\) and \(x = \pm i\).
The zeros found represent the values of \(x\) that satisfy the original polynomial equation \(f(x)=0\). Finding these zeros allows us to write the polynomial as a product of its linear factors: \(f(x) = (x-6i)(x+6i)(x-i)(x+i)\). This factorization is crucial for deeper algebraic exploration and for understanding the behavior of polynomial functions.