Problem 40 - Precalculus : Building Concepts and Connections

Question

Find an expression for a polynomial \(p(x)\) with real coefficients that satisfies the given conditions. There may be more than one possible answer. Degree \(2 ; x=\frac{1}{2}\) and \(x=\frac{3}{4}\) are zeros

Step-by-Step Solution

Answer

The polynomial \(p(x)\) that satisfies the conditions is \(p(x) = x^2 - \frac{5}{4}x + \frac{3}{8}\).

Step 1: Set up the equation

Given the polynomial to be of degree 2 and that the zeros (roots) are \(x=\frac{1}{2}\) and \(x=\frac{3}{4}\). We can write the polynomial \(p(x)\) in the factored form: \(p(x) = a(x-\frac{1}{2})(x-\frac{3}{4})\). To keep it simple, we'll assume \(a=1\). If the exercise specified a leading coefficient other than 1, that value would be used for \(a\).

Step 2: Distribute the factors

Expand the equation obtained in the step 1: \(p(x) = (x-\frac{1}{2})(x-\frac{3}{4})\). First perform the multiplication between \(x-\frac{1}{2}\) and \(x-\frac{3}{4}\).

Step 3: Simplify the equation

Once expanded, the equation needs to be simplified to present the polynomial in the standard form \(p(x) = ax^2 + bx + c\).

Key Concepts

Factoring Polynomials

Factoring polynomials is similar to breaking down a number into its prime factors; it’s a method of transforming a polynomial into a product of simpler polynomials. When we factor a polynomial, we are looking for the values that, when multiplied back together, give us the original polynomial.
For example, suppose we have a polynomial like \( p(x) = x^2 - 5x + 6 \). To factor this, we would search for two binomials \( (x - m)(x - n) \) that produce the original quadratic when multiplied together. Here, \( p(x) \) can be factored as \( (x - 2)(x - 3) \), since \( 2*3 = 6 \) and \( -2 -3 = -5 \), matching the original polynomial's constant and linear coefficients, respectively.
It's important to recognize patterns like the difference of squares, perfect square trinomials, and the sum or difference of cubes to efficiently factor polynomials.

Polynomial Equations

Polynomial equations are mathematical expressions that equate a polynomial to another value, most commonly zero. Solving polynomial equations involves finding the values of the variable that satisfy the equation—the solutions or roots of the polynomial.
These equations can vary widely in complexity, starting from linear equations (degree 1), which are the simplest form, to higher-degree equations, such as quadratic (degree 2), cubic (degree 3), and so on. The degree of the polynomial provides vital information, such as the maximum number of roots it may have.
A straightforward example is solving a quadratic equation by factoring it into binomials and then applying the zero product property, which implies that if the product of two factors is zero, at least one of the factors must be zero. Using this property, each factor of the polynomial equation can be set equal to zero to solve for the variable.

Finding Zeros of Polynomials

Finding the zeros of a polynomial is essential because it tells us where the graph of the polynomial intersects the x-axis. The zeros of a polynomial are the solutions to the equation \( p(x) = 0 \).
For example, in the exercise given, \( p(x) \) is a degree 2 polynomial with zeros at \( x=\frac{1}{2} \) and \( x=\frac{3}{4} \). We express \( p(x) \) as \( p(x) = (x - \frac{1}{2})(x - \frac{3}{4}) \), reflecting these zeros. Through the process of expansion, or 'distributing the factors' as mentioned in the step-by-step solution, and then simplifying, we find the polynomial in standard form.
It's important to remember that the Fundamental Theorem of Algebra assures us that every non-constant single-variable polynomial with complex coefficients has at least one complex root; this includes real roots in the set of complex numbers. This theorem is a guiding principle when identifying a polynomial's roots.