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Chapter 3: Problem 92

In Exercises \(91-96,\) a polynomial function \(f(x)\) is given in both expanded and factored forms. Graph the function, and solve the equations and inequalities. Give multiplicities of solutions when applicable. \(f(x)=x^{3}+4 x^{2}-11 x-30\) \(=(x-3)(x+2)(x+5)\) (a) \(f(x)=0\) (b) \(f(x)<0\) (c) \(f(x)>0\)

Short Answer

Expert verified
Roots: \( x = -5, -2, 3 \). \( f(x) < 0 \) for \( -2 < x < 3 \). \( f(x) > 0 \) for \( x < -5, -5 < x < -2, x > 3 \). Each root has multiplicity one.

Step by step solution

01

Understand the polynomial forms

The polynomial function is given in both expanded form \(f(x) = x^3 + 4x^2 - 11x - 30\) and factored form \(f(x) = (x - 3)(x + 2)(x + 5)\). Recognize that the factored form is useful for finding the roots (solutions) of the polynomial.
02

Find the roots of the polynomial

Set the factored form of the polynomial equal to zero: \( (x - 3)(x + 2)(x + 5) = 0 \). Solve for the values of x: \( x = 3, x = -2, x = -5 \). These are the roots where the polynomial equals zero.
03

Determine the multiplicities of the roots

Since the factored form \( (x - 3)(x + 2)(x + 5) \) shows each factor to the power of one, each root \( x = 3, x = -2, x = -5 \) has a multiplicity of one.
04

Solve the inequality \(f(x) < 0\)

To determine where the polynomial is less than zero, establish the intervals set by the roots: \( (-\infty, -5) \), \( (-5, -2) \), \( (-2, 3) \), \( (3, \infty) \). Test sample points within these intervals to check the sign of the polynomial in each interval: For \(x < -5\), choose \( x = -6 \): \( (-6-3)(-6+2)(-6+5) = (-9)(-4)(-1) > 0 \)For \(-5 < x < -2\), choose \( x = -4 \): \( (-4-3)(-4+2)(-4+5) = (-7)(-2)(1) > 0 \)For \(-2 < x < 3\), choose \( x = 0 \): \( (0-3)(0+2)(0+5) = (-3)(2)(5) < 0 \)For \(x > 3\), choose \( x = 4 \): \( (4-3)(4+2)(4+5) = (1)(6)(9) > 0 \)Thus, \( f(x) < 0 \) for \( -2 < x < 3 \).
05

Solve the inequality \(f(x) > 0\)

From the intervals checked: For \(f(x) > 0 \): \( x < -5 \), \( -5 < x < -2 \), and \( x > 3 \). Thus, \( f(x) > 0 \) for \( x \in (-\infty, -5) \cup (-5, -2) \cup (3, \infty) \).
06

Graph the polynomial function

Plot the roots \( x = -5, -2, 3 \) on a coordinate plane. Mark points where the polynomial changes sign according to intervals examined. Make sure the graph crosses the x-axis at the roots.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factored Form
The factored form of a polynomial is a way of expressing the polynomial as a product of its factors. This form is extremely useful because it easily reveals the roots of the polynomial.
For instance, the factored form of the polynomial provided in the exercise is \((x - 3)(x + 2)(x + 5)\). This indicates that the polynomial can be broken into three factors: \(x - 3\), \(x + 2\), and \(x + 5\).
The factored form makes it easy to find the polynomial's roots, the points where the polynomial equation equals zero. Using the given factored form, setting each factor equal to zero gives us:
  • \(x - 3 = 0 \Rightarrow x = 3\)
  • \(x + 2 = 0 \Rightarrow x = -2\)
  • \(x + 5 = 0 \Rightarrow x = -5\)
These values of \(x\) are the roots of the polynomial.
Roots of Polynomial
The roots of a polynomial are the values of \(x\) at which the polynomial equals zero. These roots are found using the factored form of the polynomial.
Given the polynomial \(f(x) = x^3 + 4x^2 - 11x - 30\), its factored form is \((x - 3)(x + 2)(x + 5)\).
The roots are determined by setting each factor equal to zero:
  • \(x - 3 = 0 \Rightarrow x = 3\)
  • \(x + 2 = 0 \Rightarrow x = -2\)
  • \(x + 5 = 0 \Rightarrow x = -5\)
Thus, the roots are \(x = 3, x = -2\), and \(x = -5 \). These are the points where the graph of the polynomial touches or crosses the x-axis.
Solving Inequalities
To solve polynomial inequalities, we first need to identify the regions where the polynomial is positive or negative.
Using the roots found earlier (\(x = 3, x = -2, x = -5\)), the real number line is divided into intervals. The intervals are \((-\infty, -5)\), \((-5, -2)\), \((-2, 3)\), and \((3, \infty)\).
We test sample points from each interval to determine whether the polynomial's value in that interval is positive or negative:
  • For \(x < -5\), picking \(x = -6\) gives \(f(-6) = (-9)(-4)(-1) > 0\).
  • For \(-5 < x < -2\), picking \(x = -4\) gives \(f(-4) = (-7)(-2)(1) > 0\).
  • For \(-2 < x < 3\), picking \(x = 0\) gives \(f(0) = (-3)(2)(5) < 0\).
  • For \(x > 3\), picking \(x = 4\) gives \(f(4) = (1)(6)(9) > 0\).
Therefore,
\( f(x) < 0 \) for \(-2 < x < 3\).
\( f(x) > 0 \) for \((-\infty, -5) \cup (-5, -2) \cup (3, \infty)\).
Multiplicity of Solutions
Multiplicity refers to the number of times a particular root appears for a given polynomial.
If a root \(x = a\) appears \(k\) times in the factored form of the polynomial, then \(a\) is said to have a multiplicity of \k\.
For example, in the polynomial \(f(x) = (x - 3)(x + 2)(x + 5)\), each factor appears only once.
Therefore, each root has a multiplicity of one:
  • \(x = 3\)
  • \(x = -2\)
  • \(x = -5\)
The multiplicity of a root affects the graph of the polynomial. If a root has even multiplicity, the graph touches the x-axis and turns around.
If a root has odd multiplicity, the graph crosses the x-axis. In this case, all roots cross the x-axis because all roots have a multiplicity of one.

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