Problem 32
Question
Find the zeros of each function. State the multiplicity of multiple zeros. $$ y=3 x^{3}-3 x $$
Step-by-Step Solution
Verified Answer
The zeros of the function \(y = 3x^3 - 3x\) are \(x = -1\), \(x = 0\), and \(x = 1\) each with multiplicity 1.
1Step 1 - Factor Out the Common x
Begin by identifying the common factor in the terms of the polynomial. Here, each term contains an x, so you can factor out an x to simplify the expression: \( y = x(3x^2 - 3) \).
2Step 2 - Factor the Quadratic
Factor the remaining quadratic expression \(3x^2 - 3\). It can be factored by taking out the common factor of 3: \( y = x(3(x^2 - 1)) \).
3Step 3 - Factor Further Using Difference of Squares
The quadratic expression \(x^2 - 1\) is a difference of squares and can be factored further using the identity \(a^2 - b^2 = (a+b)(a-b)\). Hence, it can be factored into \((x+1)(x-1)\): \( y = x(3)(x+1)(x-1) \).
4Step 4 - Find the Zeros of the Function
Set each factor equal to zero and solve for x: \(x = 0\), \(x+1 = 0\), and \(x-1 = 0\). The solutions are \(x = 0\), \(x = -1\), and \(x = 1\).
5Step 5 - Determine the Multiplicity
In this case, each zero occurs only once in the factored form of the polynomial, so each zero has a multiplicity of 1.
Key Concepts
Factoring PolynomialsDifference of SquaresPolynomial Roots Multiplicity
Factoring Polynomials
The process of factoring polynomials is essential in finding the zeros of a polynomial function, which are the values of x that make the polynomial equal to zero. These solutions are also known as the 'roots' or 'solutions' of the equation. Factoring involves breaking down a complex expression into simpler ones, products of factors, that when multiplied together give back the original polynomial.
As in our example, the first step typically involves factoring out the greatest common factor that appears in each term of the polynomial. In this case, each term of the polynomial shared a factor of x, leading to a simplified expression. After factoring out the common factor, the polynomial may reduce to a quadratic or another simpler form that can be factored further.
It's crucial to recognize different types of expressions that can be factored, such as the difference of squares which appears after the common factor is factored out. Factoring is a foundational skill for algebra students and understanding it will make finding zeros much more straightforward.
As in our example, the first step typically involves factoring out the greatest common factor that appears in each term of the polynomial. In this case, each term of the polynomial shared a factor of x, leading to a simplified expression. After factoring out the common factor, the polynomial may reduce to a quadratic or another simpler form that can be factored further.
It's crucial to recognize different types of expressions that can be factored, such as the difference of squares which appears after the common factor is factored out. Factoring is a foundational skill for algebra students and understanding it will make finding zeros much more straightforward.
Difference of Squares
The difference of squares is a specific type of polynomial that can be readily factored, given the form \(a^2 - b^2\) where both 'a' and 'b' are real numbers or algebraic expressions. The unique identity for the difference of squares is \(a^2 - b^2 = (a+b)(a-b)\).
This identity allows you to rewrite the difference of two perfect squares as the product of two factors. For example, in the polynomial \(x^2 - 1\), which appears after factoring out the common factor of 3 in our exercise, it can be recognized as \(x^2 - 1^2\) and factored into \(x+1)(x-1)\).
Knowing the difference of squares formula is incredibly useful in algebra, as it frequently appears in quadratic expressions and helps simplify complex polynomials into forms whose zeros can be easily determined.
This identity allows you to rewrite the difference of two perfect squares as the product of two factors. For example, in the polynomial \(x^2 - 1\), which appears after factoring out the common factor of 3 in our exercise, it can be recognized as \(x^2 - 1^2\) and factored into \(x+1)(x-1)\).
Knowing the difference of squares formula is incredibly useful in algebra, as it frequently appears in quadratic expressions and helps simplify complex polynomials into forms whose zeros can be easily determined.
Polynomial Roots Multiplicity
Polynomial roots multiplicity refers to the number of times a particular root appears as a factor of a polynomial. A root's multiplicity affects the shape of the polynomial's graph at the interception point with the x-axis.
For instance, a root with a multiplicity of 1, also known as a simple root, implies that the graph of the polynomial crosses the x-axis at this point. If a root has an even multiplicity, the graph of the polynomial will touch and bounce off the x-axis. For an odd multiplicity higher than 1, the graph still crosses the x-axis but with a flatter approach or departure.
In the provided exercise, the polynomial factors into \(x(3)(x+1)(x-1)\), and we set each factor equal to zero to find the roots. Each root here has multiplicity 1, indicating they are simple roots where their corresponding graph will intersect the x-axis at three distinct points. Understanding root multiplicity not only assists in solving polynomials but also in predicting the behavior of their graphs.
For instance, a root with a multiplicity of 1, also known as a simple root, implies that the graph of the polynomial crosses the x-axis at this point. If a root has an even multiplicity, the graph of the polynomial will touch and bounce off the x-axis. For an odd multiplicity higher than 1, the graph still crosses the x-axis but with a flatter approach or departure.
In the provided exercise, the polynomial factors into \(x(3)(x+1)(x-1)\), and we set each factor equal to zero to find the roots. Each root here has multiplicity 1, indicating they are simple roots where their corresponding graph will intersect the x-axis at three distinct points. Understanding root multiplicity not only assists in solving polynomials but also in predicting the behavior of their graphs.
Other exercises in this chapter
Problem 32
A salad bar offers eight choices of toppings for lettuce. In how many ways can you choose four or five toppings?
View solution Problem 32
Which number is a root of \(f(x)=x^{3}+6 x^{2}+9 x\) that has multiplicity 1\(?\) $$ \begin{array}{lllll}{\text { A. } 3} & {\text { B. } 1} & {\text { C. } 0}
View solution Problem 32
Solve each equation. $$ x^{4}+8 x^{2}+15=0 $$
View solution Problem 32
Use synthetic division and the Remainder Theorem to find \(P(a)\). $$P(x)=2 x^{3}+4 x^{2}-10 x-9 ; a=3$$
View solution