Problem 42 - Precalculus: Functions and Graphs

Question

Show that \(x-c\) is not a factor of \(f(x)\) for any real number \(c\). $$f(x)=-x^{4}-3 x^{2}-2$$

Step-by-Step Solution

Answer

No real \(c\) makes \(f(c)=0\); hence, \(x-c\) is never a factor.

Step 1: Understand the Factor Theorem

The Factor Theorem states that \(x-c\) is a factor of \(f(x)\) if and only if \(f(c) = 0\). To determine if \(x-c\) is not a factor for any real number \(c\), we must show that there is no real \(c\) such that \(f(c) = 0\).

Step 2: Analyze the Given Polynomial Function

The given polynomial function is \(f(x) = -x^4 - 3x^2 - 2\). Notice that all terms in this polynomial have negative coefficients.

Step 3: Determine if There Are Real Roots by Checking Simplified Form

Assume \(f(c) = 0\) for some real number \(c\). This implies \(-c^4 - 3c^2 - 2 = 0\). Rearrange this to give \(c^4 + 3c^2 + 2 = 0\), which must hold for \(c\).

Step 4: Investigate the Nature of the Function for Real Roots

The form \(c^4 + 3c^2 + 2\) is the sum of a quartic and a quadratic function, both of which result in non-negative values for any real \(c\), not allowing the sum to be \(0\). Break down: 1. \(c^4 \geq 0\) for all \(c\). 2. \(3c^2 \geq 0\) for all \(c\). 3. Adding \(2\) ensures that the left-hand side is always positive.

Step 5: Conclude the Non-existence of Real Roots

Since the expression \(c^4 + 3c^2 + 2\) is always greater than zero for any real \(c\), \(f(c)\) is also always negative, implying no real number \(c\) exists for which \(f(c)=0\).

Key Concepts

Factor Theorem

The Factor Theorem is a powerful tool in understanding polynomials. It tells us that a polynomial, say \(f(x)\), has a factor \(x-c\) if substituting \(c\) into the function results in zero. In simpler terms, \(x-c\) is a factor of \(f(x)\) if \(f(c) = 0\).
This theorem is very useful because it connects the concepts of factors and roots of polynomials. If we can find such a \(c\), we say \(c\) is a root or zero of the polynomial.
To apply the Factor Theorem to the given polynomial \(f(x) = -x^4 - 3x^2 - 2\), imagine substituting any real number \(c\) into \(f(x)\). We are looking to see if any \(f(c) = 0\). If no such \(c\) exists, then there is no factor \(x-c\). This breaks down the search for factors into a simple problem of substitution and checking for zero value.

Roots of Polynomial Equations

Finding the roots of a polynomial equation means finding solutions where the polynomial equals zero. For a polynomial like \(f(x) = -x^4 - 3x^2 - 2\), it's about finding values of \(x\) for which \(f(x) = 0\). These roots tell us important things about the polynomial, such as how the graph of the polynomial behaves and where it crosses the x-axis.
However, the task is to show that no real number \(c\) makes \(f(x)\) zero. By rewriting the equation as \(c^4 + 3c^2 + 2 = 0\) and analyzing it, we see the challenge: the left side is always positive or zero for any real \(c\).
Each part—\(c^4\) and \(3c^2\)—is non-negative, and adding 2 makes sure the whole expression is always positive. This means the polynomial can't equal zero for any real \(c\), indicating that it has no real roots.

Real Roots

Real roots are values of \(x\) that make a polynomial equation equal to zero, where the solutions are real numbers rather than complex or imaginary. Checking for real roots in \(f(x) = -x^4 - 3x^2 - 2\) requires examining the expression more closely.
We want to solve \(c^4 + 3c^2 + 2 = 0\) to see if any real numbers can satisfy this. Since both \(c^4\) and \(3c^2\) are always zero or positive for real numbers, the smallest possible value the expression can take is 2, when \(c^4 = 0\) and \(3c^2 = 0\).
This means the equation never reaches zero for any real \(c\), as it's impossible for a sum that includes a non-zero constant to be zero. Thus, no real roots exist for this polynomial, confirmed by the constant positive nature of \(c^4 + 3c^2 + 2\).