Problem 21 - Precalculus: Functions and Graphs

Question

Find the zeros of \(f(x),\) and state the multiplicity of each zero. $$f(x)=x^{4}+7 x^{2}-144$$

Step-by-Step Solution

Answer

The zeros of \(f(x)\) are 3 and -3, each with multiplicity 1.

Step 1: Set the Function to Zero

To find the zeros of the function, set the function equal to zero: \[ f(x) = x^4 + 7x^2 - 144 = 0 \]

Step 2: Substitution for Simplification

Let's substitute \( y = x^2 \). Therefore, the equation becomes: \[ y^2 + 7y - 144 = 0 \] This reduces our fourth-degree polynomial to a quadratic equation in terms of \( y \).

Step 3: Apply the Quadratic Formula

Apply the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 7, c = -144 \): \[ y = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-144)}}{2 \times 1} \]\[ y = \frac{-7 \pm \sqrt{49 + 576}}{2} \]\[ y = \frac{-7 \pm \sqrt{625}}{2} \]\[ y = \frac{-7 \pm 25}{2} \]

Step 4: Solve for y

Calculate the solutions for \( y \):\( y_1 = \frac{-7 + 25}{2} = 9 \)\( y_2 = \frac{-7 - 25}{2} = -16 \)

Step 5: Back-Substitute to x

Substitute back \( y = x^2 \) to find \( x \): For \( y_1 = 9 \):\( x^2 = 9 \)\( x = \pm 3 \) For \( y_2 = -16 \): Since the equation \( x^2 = -16 \) has no real solutions (as squares of real numbers are non-negative), we disregard it for real zeros.

Step 6: Determine Zero Multiplicities

The solutions found were \( x = 3 \) and \( x = -3 \). Each corresponds to a unique result of the quadratic factorization and each zero appears once, so their multiplicities are both 1.

Key Concepts

Quadratic Substitution

Quadratic substitution is a handy technique used to simplify higher degree polynomials, allowing us to use methods designed for quadratic equations. In this particular exercise, we replace the expression involving a quadratic power, like with the original polynomial equation

• \( x^4 + 7x^2 - 144 = 0 \)

with a simpler quadratic form.
To achieve this, set:

• \( y = x^2 \)

Transforming the equation to:

• \( y^2 + 7y - 144 = 0 \)

This substitution reduces the complexity, converting the fourth-degree polynomial into a well-known quadratic form. We can now proceed by using standard methods to find solutions for this simpler equation.
After solving for \( y \), remember to substitute back to find the original variable \( x \), ensuring you cover both positive and negative square roots.

Zero Multiplicity

Multiplicity of a zero refers to how many times a particular solution to a polynomial equation appears. It's essential for understanding polynomial behavior at its roots. In the original problem, we seek the zeros of

• \( x = 3 \)
• \( x = -3 \)

Each solution from the quadratic substitution corresponds to a distinct zero of the polynomial function.
To determine the multiplicity, observe the fact that each zero appears only once after solving the polynomial equation. Thus, the multiplicities of both \( x = 3 \) and \( x = -3 \) are 1. A multiplicity higher than 1 would imply that the polynomial equation could be factored further to repeat that zero. Such repetitions indicate that the function merely "touches" the x-axis rather than crossing it.

Quadratic Formula

The quadratic formula is a universal tool used to solve any quadratic equation of the form

• \( ax^2 + bx + c = 0 \)

It provides solutions through:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In this exercise, once we performed quadratic substitution, we used the quadratic formula to solve

• \( y^2 + 7y - 144 = 0 \)

Plugging in the coefficients \( a = 1, b = 7, c = -144 \), we computed:

• \( y = \frac{-7 \pm \sqrt{625}}{2} \)

yielding solutions \( y = 9 \) and \( y = -16 \). This calculation highlights the efficiency of the quadratic formula in identifying zero points, especially when the polynomial cannot be factored easily.

Real and Complex Solutions

Polynomials can have both real and complex solutions, depending on the discriminant

• \( b^2 - 4ac \)

from the quadratic formula. Importantly, real solutions occur when the discriminant is non-negative, while negative results yield complex solutions. Here,

• \( y = x^2 = 9 \)

provides real solutions since the square root is positive.
The negative square root

• \( x^2 = -16 \)

suggests complex solutions \( x = \pm 4i \), as no real number squared equals a negative value.
In practical applications, if only real zeros are needed, negative discriminants are typically disregarded. Understanding both real and complex solutions helps in gaining a comprehensive view of a polynomial's behavior and the nature of its roots.