Problem 13 - Precalculus : Building Concepts and Connections

Question

Find all the zeros, real and nonreal, of the polynomial. Then express \(p(x)\) as a product of linear factors. $$p(x)=x^{3}+5 x$$

Step-by-Step Solution

Answer

The zeros of the polynomial \(x^{3}+5x = 0\) are \(0, i√5, -i√5\), and the polynomial expressed as a product of linear factors is \(x(x - i√5)(x + i√5)\).

Step 1: Set the polynomial equal to zero

We begin this process by setting the polynomial \(p(x)\) equal to zero to enable us to find the values of x that satisfy this requirement. So: \(x^{3}+5x = 0\)

Step 2: Factor out common terms

Next, we factor out any common factors from each term in the polynomial. In this case, x is a common factor, so: \(x(x^{2}+5) = 0\)

Step 3: Set each factor equal to zero

Now we set each factor equal to zero and solve for x. From \(x = 0\) we get our first root which is zero, and from \(x^{2}+5 = 0\), we get two complex factors by solving \(x^{2} = -5\), namely \(x = +√(-5) = +i√5\) and \(x = -√(-5) = -i√5\)

Step 4: Express the Polynomial as a Product of Linear Factors

Our final step is to express \(p(x)\) as a product of linear factors with the help of the zeros we have found. This gives: \(p(x)=x(x - i√5)(x + i√5)\)

Key Concepts

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial into simpler terms or expressions known as factors, which when multiplied together yield the original polynomial. It is akin to finding numbers that multiply to give another number, like factors of 12 are 3 and 4, because 3 times 4 equals 12. In math terms, if you have a polynomial like

• \(x^3 + 5x\)

you can factor it by identifying the common terms in each part of the polynomial.

In our example, we can see that \(x\) is a common factor in both terms \(x^3\) and \(5x\). Factoring \(x\) out, the expression becomes:

• \(x(x^2 + 5)\)

To completely factor polynomials, you may need further steps if the remaining expressions can be broken down more. Factoring is essential for simplifying expressions and solving equations, especially when trying to find the zeros or roots of equations.

Linear Factors

Linear factors are expressions of the form \(ax + b\), where the variable \(x\) is raised to the first power, making them 'linear'. When you express a polynomial as a product of linear factors, you break it down into pieces that are multiplied to give you the polynomial.

For example, if we take the polynomial:

• \(p(x) = x^3 + 5x\)

After factoring out the common term \(x\), we solve the simpler equation \(x(x^2 + 5) = 0\) as:

• \(x = 0\)
• \(x^2 + 5 = 0\)

This leads us to roots, also known as zeros, of the polynomial.

Each root contributes to a linear factor: if \(x = a\) is a root, a factor is \((x - a)\). Thus in the polynomial above, the linear factors are:

• \(x\)
• \((x - i\sqrt{5})\)
• \((x + i\sqrt{5})\)

These factors reveal the polynomial's structure and connection to its zeros.

Complex Numbers

Complex numbers extend our number system to better handle equations with square roots of negative numbers. They have the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, satisfying the property \(i^2 = -1\).

When solving the quadratic equation \(x^2 + 5 = 0\), we encounter a case where the solution involves taking a square root of a negative number.

• \(x^2 = -5\)

This prompts the necessity of complex numbers, because the solution becomes

• \(x = \pm i\sqrt{5}\)

Here, \(i\sqrt{5}\) represents a complex number.

Utilizing complex numbers allows us to fully capture the behavior of polynomials, revealing not only real roots but also complex ones. This holistic approach is why mathematics relies on complex numbers, particularly when working with linear factors of polynomials with no real solutions.