Problem 62 - Precalculus Mathematics for Calculus

Question

Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(a)\) $$P(x)=-x^{4}+10 x^{2}+8 x-8$$

Step-by-Step Solution

Answer

The real zeros are \(\sqrt{5 - \sqrt{33}}, -\sqrt{5 - \sqrt{33}}, \sqrt{5 + \sqrt{33}}, -\sqrt{5 + \sqrt{33}}\).

Step 1: Identify Potential Real Zeros

First, identify the polynomial's degree, which is 4, and recognize that it may have up to four real zeros. The polynomial can be written as: \(P(x) = -x^{4} + 10x^{2} + 8x - 8\).

Step 2: Substitution for Quadratic Form

Notice that the polynomial is in quadratic form with \(-x^4 + 10x^2\), which suggests substituting \(y = x^2\) into the polynomial to make it easier to solve. Substitute to get \(Q(y) = -y^2 + 10y + 8\).

Step 3: Apply the Quadratic Formula

Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) on \(Q(y) = -y^2 + 10y + 8\), where \(a = -1\), \(b = 10\), and \(c = 8\). Substitute these values:\[y = \frac{-10 \pm \sqrt{10^2 - 4(-1)(8)}}{2(-1)} = \frac{-10 \pm \sqrt{100 + 32}}{-2}\]

Step 4: Simplify the Quadratic Formula

Calculate under the square root: \(\sqrt{132} = \sqrt{4 \times 33}\), simplify to get \(2\sqrt{33}\). Thus,\[ y = \frac{-10 \pm 2\sqrt{33}}{-2}\] gives possible solutions:\[y_1 = 5 - \sqrt{33},\ y_2 = 5 + \sqrt{33}\]

Step 5: Find Real Zeros of Original Polynomial

To find the real zeros of the original polynomial, use the solutions for \(y = x^2\), with \(y_1 = 5 - \sqrt{33}\) and \(y_2 = 5 + \sqrt{33}\). Solve for \(x\) in each case:For \(y_1 = x^2\):\[ x = \pm \sqrt{5 - \sqrt{33}} \]For \(y_2 = x^2\): \[ x = \pm \sqrt{5 + \sqrt{33}} \]These give four possible real zeros for \(P(x)\).

Step 6: Conclude with Real Zeros

The final solutions for real zeros of the polynomial \(P(x)\) are:\[x = \sqrt{5 - \sqrt{33}},\ x = -\sqrt{5 - \sqrt{33}},\ x = \sqrt{5 + \sqrt{33}},\ x = -\sqrt{5 + \sqrt{33}}\]. Since all values are real, these are the zeros.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool in algebra used to find the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \). It is given by:

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

It provides a straightforward way to solve quadratic equations when factoring is complex or impossible. To apply the quadratic formula:

• Identify the coefficients \( a \), \( b \), and \( c \) in the quadratic expression.
• Substitute these values into the formula.
• Simplify under the square root to determine the discriminant \( b^2 - 4ac \), which indicates the number and type of solutions.

The discriminant is crucial:

• If it's positive, there are two real solutions.
• If it's zero, there's exactly one real solution.
• If negative, the solutions are complex, not real.

In the given problem, using the quadratic formula helped simplify finding the roots by dealing directly with the transformed quadratic expression.

Polynomial Degree

The degree of a polynomial is the highest power of the variable in the polynomial expression. In the polynomial given, \( P(x) = -x^4 + 10x^2 + 8x - 8 \), the degree is 4 because the highest power of \( x \) is 4. The degree of a polynomial gives important information about its behavior and characteristics:

• It indicates the maximum number of solutions or roots the polynomial equation can have. A fourth-degree polynomial, like this one, can have up to four real roots.
• The degree affects the end behavior of the polynomial graph. For example, because the leading term is \(-x^4\), the graph will open downwards on both ends.
• The degree also guides the methods used for solving solutions or factoring.

Understanding the degree helps anticipate the potential complexity of finding solutions and prepares us for what methods may be appropriate.

Substitution Method

The substitution method is a handy technique used to simplify complex polynomial equations by introducing a new variable. In essence, we transform the equation into a simpler form that is easier to solve. Here's how it works:

• Identify parts of the polynomial that form a recognizable or factorable expression, such as a quadratic form.
• Introduce a substitution variable, such as \( y = x^2 \), to convert these parts into a simpler quadratic equation.
• In the exercise, \( -x^4 + 10x^2 \) was simplified to \( -y^2 + 10y \) through substitution. This makes solving the equation more straightforward.

Once the simpler equation is solved, substitute back to find the solution for the original variable. This method is useful when polynomials exhibit a structure that can be restated in a simpler quadratic form, which can then be addressed with methods like factoring or using the quadratic formula.

Quartic Polynomial

A quartic polynomial is a polynomial of degree four, meaning the highest power of the variable \( x \) is four. An example is \( P(x) = -x^4 + 10x^2 + 8x - 8 \). Quartic polynomials can be tricky to deal with due to their higher degree and the potential number of real and complex roots. Key features include:

• Having up to four real roots, which can be determined using techniques such as factorization, graphing, or substitution, as seen in the problem.
• The presence of a leading negative coefficient, \(-x^4\), influences the graph's end behavior—opening downwards on both ends.
• Quartic polynomials can often be broken down through substitution into simpler polynomial forms, as demonstrated in the problem.

Understanding the structure and characteristics of quartic polynomials is essential for solving them effectively, making techniques like substitution and the quadratic formula invaluable in tackling such equations.