Problem 16 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the (real) zeros of the polynomial given. (a) \(f(x)=-x^{3}-x^{2}-5 x\) (b) \(g(x)=0.5 x^{4}-0.5\)

Step-by-Step Solution

Answer

The real zeroes of \(f(x)\) are 0 and \(f(x)\) has no other real solutions. The real zeroes of \(g(x)\) are 1 and -1, and \(g(x)\) also has no other real solutions.

Step 1: Preparation

Analyze the polynomials and, if possible, factor out any common factors.

Step 2: Solving for Polynomial (a)

First equation is \(f(x) = -x^{3} - x^{2} - 5x\). Start by factoring out the common factor \(x\), to get \(f(x) = x(-x^{2} - x - 5)\). Now, set \(f(x)\) equal to zero and solve for \(x\). This gives the solutions \(x=0\) and \(x\) such that \(-x^{2} - x - 5 = 0\). By solving the quadratic equation, no real solutions can be found.

Step 3: Solving Polynomial (b)

Second equation is \(g(x) = 0.5x^{4} - 0.5\). Set \(g(x)\) equal to zero and solve for \(x\). This gives the equation \(0.5x^{4} - 0.5 = 0\). By multiplying everything by 2 to remove the decimal, the equation becomes \(x^{4} - 1 = 0\). That's the difference of squares, so it can be factored into \((x^{2} - 1)(x^{2} + 1) = 0\). Continuing to factor, we get \((x - 1)(x + 1)(x^{2} + 1) = 0\). Solving each part equals to zero results in the solutions: \(x = 1\), \(x = -1\) and no real solutions from \(x^{2} + 1 = 0\).

Key Concepts

Polynomial Factorization

The process of breaking down a polynomial into simpler components, which when multiplied together give back the original polynomial, is known as polynomial factorization. It's akin to dismantling a complex piece of machinery into its basic parts to understand how it functions. Polynomial factorization dramatically simplifies the task of finding zeros, which are the points at which the polynomial touches or crosses the x-axis on a graph.

For example, consider the polynomial from exercise (a), \(f(x) = -x^3 - x^2 - 5x\). Factoring out the greatest common factor (GCF), in this case \(x\), is the first step. We get \(f(x) = x(-x^2 - x - 5)\), which reveals the first zero as \(x=0\). The remaining quadratic equation, \( -x^2 - x - 5\), can be further investigated for possible zeros through techniques such as the quadratic formula, completing the square, or factoring, if applicable.

Quadratic Equations

Quadratic equations are second-degree polynomials typically in the form \(ax^2 + bx + c = 0\). These equations are quintessential in algebra and have diverse applications across mathematics. The solutions to these equations, where the graph intersects the x-axis, are known as the 'roots' or 'zeros' of the equation.

In the context of our exercise (a), once we factored out the GCF, we were left with a quadratic equation \( -x^2 - x - 5\) that didn't yield real solutions. This lack of real solutions indicates that the parabola represented by the quadratic equation does not touch the x-axis. In such a case, this portion of the polynomial only contributes complex zeroes, which fall outside the realm of 'real' solutions.

Difference of Squares

The 'difference of squares' is a special factoring pattern where a two-term polynomial in the form \(a^2 - b^2\) can be broken down into \( (a + b)(a - b)\). Recognizing this pattern is crucial when manipulating algebraic expressions and solving polynomial equations.

Consider the polynomial from exercise (b), \(g(x) = 0.5x^4 - 0.5\). Upon multiplying by 2, we get \(x^4 - 1\), which is a clear example of the difference of squares pattern. Here, \(a = x^2\) and \(b = 1\), allowing us to factor it into \( (x^2 - 1)(x^2 + 1)\). This further factors down as \( (x - 1)(x + 1)(x^2 + 1)\), revealing two real zeros at \(x = 1\) and \(x = -1\), whereas \(x^2 + 1\) does not yield any real solutions since it would require the square root of a negative number.