Problem 82 - Algebra and Trigonometry

Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam.Find the real zeros of $$ h(x)=\frac{\left(x^{4}+1\right) \cdot 2 x-\left(x^{2}-1\right) \cdot 4 x^{3}}{\left(x^{4}+1\right)} $$

Step-by-Step Solution

Answer

The real zeros of the function are 0, \( \pm \sqrt{1+\sqrt{2}} \), \( \pm \sqrt{1-\sqrt{2}} \).

Step 1: Simplify the function

First, simplify the given function. The function provided is \[ h(x) = \frac{(x^4 + 1) \cdot 2x - (x^2 - 1) \cdot 4x^3}{x^4 + 1} \]Expand and simplify the numerator:

Step 2: Expand the numerator

Distribute the terms in the numerator:\[ (x^4 + 1) \cdot 2x = 2x^5 + 2x \]\[ (x^2 - 1) \cdot 4x^3 = 4x^5 - 4x^3 \]So the numerator becomes:\[ 2x^5 + 2x - 4x^5 + 4x^3 \]

Step 3: Combine like terms

Combine the like terms in the numerator:\[ (2x^5 - 4x^5) + 4x^3 + 2x = -2x^5 + 4x^3 + 2x \]

Step 4: Substitute back into the function

Substitute the simplified numerator back into the original function:\[ h(x) = \frac{-2x^5 + 4x^3 + 2x}{x^4 + 1} \]

Step 5: Find the real zeros of the simplified function

The real zeros of \( h(x) \) are the values of \( x \) such that the numerator equals zero while the denominator is non-zero. Set the numerator equal to zero:\[ -2x^5 + 4x^3 + 2x = 0 \]Factor out the common term, \( 2x \):\[ 2x(-x^4 + 2x^2 + 1) = 0 \]This gives us two equations to solve:\( 2x = 0 \) and \( -x^4 + 2x^2 + 1 = 0 \).

Step 6: Solve for the first factor

Solve the first factor:\[ 2x = 0 \]\[ x = 0 \]

Step 7: Solve for the second factor

Solve the second factor, \( -x^4 + 2x^2 + 1 = 0 \). Let \( y = x^2 \), then the equation becomes: \[ -y^2 + 2y + 1 = 0 \]Solve this quadratic equation using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = -1, b = 2, c = 1 \):\[ y = \frac{-2 \pm \sqrt{2^2 - 4(-1)(1)}}{2(-1)} = \frac{-2 \pm \sqrt{4 + 4}}{-2} = \frac{-2 \pm 2\sqrt{2}}{-2} = 1 \mp \sqrt{2} \]Reverting back to \( x^2 \): \[ x^2 = 1 + \sqrt{2} \text{ or } x^2 = 1 - \sqrt{2} \]The real solutions for \( x \) are: \[ x = \pm \sqrt{1+\sqrt{2}} \text{ or } x = \pm \sqrt{1-\sqrt{2}} \]

Key Concepts

Polynomial Functions

Polynomial functions are expressions involving variables raised to positive integer powers. They have the general form \( f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \). These functions are fundamental in algebra and appear frequently in various mathematical problems. They can represent linear, quadratic, cubic, and higher-degree functions. In this problem, the function \( h(x) \) is a polynomial function. We need to simplify it first to find its zeros.
The given function \( h(x) = \frac{\left(x^4+1\right) \cdot 2x - \left(x^2-1\right) \cdot 4x^3}{x^4+1} \) is a rational function, but the numerator and denominator involve polynomials. Simplifying polynomial functions helps us understand and solve them efficiently by focusing on their critical parts.

Factoring

Factoring is the process of breaking down a polynomial into simpler polynomials that, when multiplied together, reproduce the original polynomial. It's crucial for solving polynomial equations, as it helps find the roots.
In our example, after simplifying \( h(x) \), we factor the numerator \( -2x^5 + 4x^3 + 2x \) to find the zeros of the function.
We factor out the common term \( 2x \):

• \( 2x(-x^4 + 2x^2 + 1) = 0 \)
• This gives us two simpler equations: \( 2x = 0 \) and \( -x^4 + 2x^2 + 1 = 0 \)

By factoring, we reduce the complexity of the equation and prepare it for solving.

Solving Equations

Solving equations involves finding the value(s) of variables that make the equation true. In polynomial functions, we solve to find the roots or zeros. These are the points where the function crosses the x-axis.
In our example, we solve two equations:

• \( 2x = 0 \) – This is easy and gives us \( x = 0 \).
• \( -x^4 + 2x^2 + 1 = 0 \) – This is a bit more complex and requires further steps.

Solving these types of equations often involves various methods, including factoring, substitution, and using formulas like the quadratic formula. The solutions to these equations tell us the positions where the polynomial function equals zero.

Quadratic Formula

The quadratic formula is essential for solving quadratic equations (equations of the form \( ax^2 + bx + c = 0 \)). It is given by:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In our problem, after substituting \( y = x^2 \), we get a quadratic equation: \( -y^2 + 2y + 1 = 0 \). We use the quadratic formula here with:

• \( a = -1 \)
• \( b = 2 \)
• \( c = 1 \)

This gives:
\( y = \frac{-2 \pm \sqrt{2^2 - 4(-1)(1)}}{2(-1)} = \frac{-2 \pm \sqrt{4 + 4}}{-2} = \frac{-2 \pm 2\sqrt{2}}{-2} = 1 \mp \sqrt{2} \)
Reverting back, we find:

• \( x^2 = 1 + \sqrt{2} \)
• \( x^2 = 1 - \sqrt{2} \)
With these solutions:\( x = \pm \sqrt{1+\sqrt{2}} \) or \( x = \pm \sqrt{1-\sqrt{2}} \).
The quadratic formula is powerful for solving equations that would be difficult to factor directly.