Problem 49 - Algebra for College Students

Question

Argue that \(f(x)=3 x^{4}+2 x^{2}+5\) has no factor of the form \(x-c\), where \(c\) is a real number.

Step-by-Step Solution

Answer

The polynomial has no real root, so it has no factor of the form \(x-c\).

Step 1: Understand the Polynomial Factor Theorem

According to the Polynomial Factor Theorem, if \( f(x) \) has a factor of the form \( x- c \), then \( f(c) = 0 \). This implies that for a polynomial to have \( x-c \) as a factor, the value of \( c \) must make the polynomial equal to zero when substituted in.

Step 2: Check if f(x) = 0 has real roots

Consider the function \( f(x) = 3x^4 + 2x^2 + 5 \). To check for factors of the form \( x-c \), solve \( f(x) = 0 \):\[ 3x^4 + 2x^2 + 5 = 0 \]This equation can be rewritten as \( 3(x^2)^2 + 2x^2 + 5 = 0 \). This is a quadratic equation in terms of \( y = x^2 \), which gives:\[ 3y^2 + 2y + 5 = 0 \]

Step 3: Determine the discriminant

To determine if the equation \( 3y^2 + 2y + 5 = 0 \) has real solutions, compute the discriminant:\[ \Delta = b^2 - 4ac = 2^2 - 4 \times 3 \times 5 \]\[ \Delta = 4 - 60 = -56 \]

Step 4: Analyze the discriminant value

Since the discriminant \( \Delta = -56 \) is negative, the quadratic equation \( 3y^2 + 2y + 5 = 0 \) has no real solutions. This means \( f(x) = 0 \) has no real roots, and thus has no factor of the form \( x-c \) with \( c \) as a real number.

Key Concepts

Real Roots

Real roots of a polynomial are the values of the variable that solve the equation when the polynomial is set to zero. For a polynomial function like \( f(x) = 3x^4 + 2x^2 + 5 \), finding real roots means identifying the values of \( x \) that make \( f(x) = 0 \). A real root corresponds to a point where the graph of \( f(x) \) crosses or touches the x-axis.

Finding a solution involves looking for numbers \( x \) such that substituting them into \( f(x) \) results in a value of zero. If such values exist among the real numbers, we refer to them as real roots. However, in this specific problem involving a fourth-degree polynomial, identifying real roots becomes challenging because complex roots might exist, depending on the polynomial's characteristics.

In our given polynomial, it turns out there are no real roots because the related quadratic equation formed, in this case, has a discriminant that is negative, indicating no intersections with the x-axis. Therefore, no real number \( c \) makes \( f(c) = 0 \).

Quadratic Equation

Quadratic equations take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. They are polynomials of degree two, which means the highest power of the variable is squared. Quadratic equations are essential because they offer a manageable complexity while allowing us to solve a wide range of problems.

In the exercise, the polynomial \( f(x) = 3x^4 + 2x^2 + 5 \) is transformed into a quadratic equation for easier analysis. By letting \( y = x^2 \), we rewrite it as \( 3y^2 + 2y + 5 = 0 \). This substitution simplifies the original polynomial to a new form that we can analyze using quadratic techniques. Quadratic equations are key in many mathematical fields for modeling real-world situations and solving mathematical problems efficiently. In particular, they provide insights into the nature and number of the roots of higher-degree polynomials like the one in the problem.

Discriminant

The discriminant is a crucial part of understanding the nature of the roots of a quadratic equation \( ax^2 + bx + c = 0 \). It is given by the formula \( \Delta = b^2 - 4ac \). The value of the discriminant can determine whether the roots of the quadratic equation are real or not.

Here's how the discriminant works:

• If \( \Delta > 0 \), the equation has two distinct real roots.
• If \( \Delta = 0 \), the equation has exactly one real root (repeated root).
• If \( \Delta < 0 \), the equation has no real roots; instead, it has two complex roots.

In this exercise, we calculated the discriminant of the equation \( 3y^2 + 2y + 5 = 0 \). The result was \( \Delta = -56 \), which is negative. This means that the quadratic equation does not have real roots. Hence, the polynomial \( f(x) = 3x^4 + 2x^2 + 5 \) cannot be factored into the linear factor \( x - c \) with \( c \) being a real number.