Problem 66 - Algebra 2

Question

What are the zeros of the polynomial function \(y=(x-3)(2 x+1)(x-1) ?\) F. \(\frac{1}{2}, 1,3\) G. \(-1,1,3\) H. \(-\frac{1}{2}, 1,3\) J. \(-3, \frac{1}{2}, 1\)

Step-by-Step Solution

Answer

The zeros of the polynomial function are \( -\frac{1}{2}, 1, 3 \) (Option H).

Step 1: Identify the Factors of the Polynomial

The polynomial is already factored into the form \(y=(x-a)(bx+c)(x-d)\). Where, \(a=3\), \(b=2\), \(c=1\), and \(d=1\).

Step 2: Set Each Factor Equal to Zero

For a polynomial to be zero, one of the factors must be zero. Set each factor equal to zero: \((x-3)=0\), \((2x+1)=0\), and \((x-1)=0\).

Step 3: Solve for x in Each Equation

Solve each equation from Step 2 for x. \((x-3)=0\) becomes \(x=3\), \((2x+1)=0\) becomes \(2x=-1\) and after dividing by 2, \(x=-\frac{1}{2}\), and \((x-1)=0\) becomes \(x=1\).

Step 4: Combine the Zero Values

The solutions from Step 3 are the zeros of the polynomial function. Combine them into a set: \(x=\left\{-\frac{1}{2}, 1, 3\right\}\).

Key Concepts

Polynomial Equations

Polynomial equations are algebraic expressions that involve terms with exponents that are whole numbers. The highest exponent within the polynomial indicates its degree, which tells us the number of solutions, or 'zeros', it can have. For instance, a quadratic equation, which is a second-degree polynomial, can have two zeros. The zeros of a polynomial are the values of the independent variable, typically x, which make the function equal to zero.

In the context of our exercise, the polynomial equation given is \(y=(x-3)(2x+1)(x-1)\). When we talk about the zeros of this polynomial, we are looking for the values of x that would make the entire equation equal to zero. The equation is already factored, which makes finding the zeros more straightforward. The degree of this polynomial is three since the highest exponent you would get after expanding the terms is three, implying that there are three zeros to find.

Factoring Polynomials

Factoring polynomials is the process of breaking down a polynomial into simpler 'factors' that, when multiplied together, produce the original polynomial. This is crucial because it simplifies the task of finding the zeros of the polynomial. The key to factoring is identifying patterns and using various strategies like factoring out the greatest common factor, utilizing special product formulas, or applying methods like grouping.

In our exercise, the polynomial is already neatly factored into \(y=(x-3)(2x+1)(x-1)\). Each of these factors can be set to zero to find the solutions. It is important for students to learn how to both factor a given polynomial and work with already factored forms, as this is a fundamental skill in solving algebraic equations. Recognizing that the provided polynomial is factored already saves time and allows the student to proceed directly to finding the solutions.

Solving Algebraic Equations

Solving algebraic equations involves finding the value or values of the variable that make the equation true. When given a factored polynomial, the process involves setting each factor equal to zero and solving each resulting equation for the variable. This is based on the zero-product property which states that if the product of several factors is zero, at least one of the factors must be zero.

In the exercise at hand, once the factors \(x-3\), \(2x+1\), and \(x-1\) are set to zero, we solve each equation separately. This step is fundamental, and careful arithmetic ensures that each zero is found correctly. Combining the zeros found from each factor, we obtain the set of all zeros of the polynomial. Clear and deliberate steps in solving these equations help students understand the processes involved and build their problem-solving skills.