/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Factor the polynomial completely... [FREE SOLUTION] | 91Ó°ÊÓ

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Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. \(P(x)=x^{6}+16 x^{3}+64\)

Short Answer

Expert verified
The zeros are \( x = -2 \) with multiplicity 2, and \( x = 1 \pm i\sqrt{3} \) each with multiplicity 2.

Step by step solution

01

Identify the Structure

Notice that the given polynomial \( P(x) = x^6 + 16x^3 + 64 \) appears to be a quadratic in form. Consider it as a polynomial of \( x^3 \), i.e., \( (x^3)^2 + 16(x^3) + 64 \). This gives it a resemblance to the standard quadratic form \( a^2 + 2ab + b^2 \), which is a perfect square trinomial.
02

Factor the Polynomial

Write the polynomial as a perfect square by recognizing it fits the form \( (a + b)^2 = a^2 + 2ab + b^2 \). Here, \( a = x^3 \) and \( b = 8 \), making \((x^3 + 8)^2\). Therefore, factor \( P(x) \) as \((x^3 + 8)(x^3 + 8)\) or \((x^3 + 8)^2\).
03

Factor Further Using Sum of Cubes

Recognize \( x^3 + 8 \) as a sum of cubes, since \( 8 = 2^3 \). The sum of cubes formula is \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) where \( a = x \) and \( b = 2 \). Thus, \( x^3 + 8 \) factors into \((x + 2)(x^2 - 2x + 4)\).
04

Expand Factoring Result

Since \( (x^3 + 8)^2 \) was expanded to \((x + 2)(x^2 - 2x + 4)(x + 2)(x^2 - 2x + 4)\), we can write it completely factored as \((x + 2)^2(x^2 - 2x + 4)^2\).
05

Find the Zeros and Their Multiplicities

The zeros of \( P(x) \) are found by setting each factor equal to zero. Solving \( x + 2 = 0 \) gives \( x = -2 \). Solving \( x^2 - 2x + 4 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), \( c = 4 \), we get complex roots \( 1 \pm i\sqrt{3} \). The zero \( x = -2 \) has a multiplicity of 2, and each complex zero \( 1 \pm i\sqrt{3} \) also has a multiplicity of 2.

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Key Concepts

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

Zeros of a Polynomial
The zeros of a polynomial are the values of the variable, often denoted as \(x\), for which the polynomial evaluates to zero. In simpler terms, they are the solutions to the equation formed when you set the polynomial equal to zero. Finding these zeros is crucial as they are often the points where the graph of the polynomial intersects the x-axis. Understanding these intersections can help in graph sketching and comprehension of the polynomial's behavior in real-world contexts. To find the zeros, you factor the polynomial as much as possible and then solve each factor as a separate equation set to zero. For instance, when we factor the polynomial \(P(x)=x^{6}+16x^{3}+64\) into \((x+2)^2(x^2 - 2x + 4)^2\), the zeros are obtained by solving \((x+2) = 0\) and \((x^2 - 2x + 4) = 0\). This reveals that \(x = -2\), and \(x = 1 \pm i\sqrt{3}\). These solutions are known as the zeros of the polynomial. Each zero carries a multiplicity, which refers to the number of times that particular zero occurs.
In this example, \(x = -2\) has a multiplicity of 2, representing two intersections or 'touch' points on the graph at \(x = -2\). Similarly, the complex zeros \(1 \pm i\sqrt{3}\) are each repeated twice due to their multiplicity of 2.
Quadratic Formula
The quadratic formula is a celebrated mathematical tool essential for solving equations of the form \(ax^2 + bx + c = 0\). It provides a straightforward method to find the zeros of any quadratic equation, which are the values of \(x\) that satisfy the equation and make it equal to zero. The formula is expressed as: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Where:
  • \(a\) is the coefficient of \(x^2\)
  • \(b\) is the coefficient of \(x\)
  • \(c\) is the constant term
In our factoring of the polynomial \(P(x)\), the factor \(x^2 - 2x + 4\) is a quadratic that doesn't factor easily with integers. Instead, we apply the quadratic formula: here, \(a = 1\), \(b = -2\), and \(c = 4\). Substituting these values, we calculate:\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(4)}}{2(1)} = \frac{2 \pm \sqrt{4 - 16}}{2}\]This simplifies to \(x = 1 \pm i\sqrt{3}\), indicating complex roots. This example illustrates that the quadratic formula not only works for real solutions but also efficiently manages complex solutions, which arise from negative discriminants where \(b^2 - 4ac < 0\).
Sum of Cubes
The sum of cubes is a specific formula used in algebra to simplify expressions of the form \(a^3 + b^3\). This formula is useful for factoring expressions that otherwise wouldn't factor neatly. It states that: \[a^3 + b^3 = (a + b)(a^2 - ab + b^2)\]In our exercise, recognizing and applying the sum of cubes was a pivotal step. The term \(x^3 + 8\) fits this formula perfectly since \(8\) equals \(2^3\). Therefore, by setting \(a = x\) and \(b = 2\), \(x^3 + 8\) becomes \((x + 2)(x^2 - 2x + 4)\).This transformation simplifies factoring significantly, breaking down complex expressions into more manageable linear and quadratic factors. The application of the sum of cubes helps uncover the zeros more efficiently, offering a pathway to compute complex solutions through simpler, reduced expressions. Understanding and applying this formula can make polynomial factoring more accessible and allow for deeper insights into solving higher-degree equations.
Multiplicity of Zeros
Multiplicity refers to the number of times a particular zero appears as a root of the polynomial. When calculated, the multiplicity of zeros can offer insightful details about how the polynomial behaves at and around specific points.For example, if a zero appears with multiplicity 1, it indicates that the graph of the polynomial crosses the x-axis at that zero. On the other hand, a multiplicity of 2 or more suggests that the graph "touches" or "bounces off" the x-axis rather than crossing it, adding subtlety to graph interpretation. In the solution to our polynomial \(P(x) = x^6 + 16x^3 + 64\), the factorization yields zeros with specific multiplicities:
  • The zero at \(x = -2\) has a multiplicity of 2. This means that the polynomial "bounces" off the x-axis at this point.
  • The complex roots at \(1 \pm i\sqrt{3}\) each have a multiplicity of 2 as well, highlighting repeated solutions even though they are not visualized on the real number line.
Understanding multiplicity aids in predicting polynomial behavior and influences how we interpret graphs of polynomial functions. Knowing about the multiplicity of zeros ensures a richer grasp of polynomial characteristics and helps mathematicians and students alike in constructing expert solutions.

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Most popular questions from this chapter

Drug Concentration After a certain drug is injected into a patient, the concentration \(c\) of the drug in the bloodstream is monitored. At time \(t \geq 0\) (in minutes since the injection), the concentration (in mg/L) is given by $$ c(t)=\frac{30 t}{t^{2}+2} $$ (a) Draw a graph of the drug concentration. (b) What eventually happens to the concentration of drug in the bloodstream?

Use a graphing device to find all real solutions of the equation, rounded to two decimal places. $$ 2 x^{3}-8 x^{2}+9 x-9=0 $$

Graph the rational function, and find all vertical asymptotes, x- and y-intercepts, and local extrema, correct to the nearest decimal. Then use long division to find a polynomial that has the same end behavior as the rational function, and graph both functions in a sufficiently large viewing rectangle to verify that the end behaviors of the polynomial and the rational function are the same. $$ r(x)=\frac{x^{4}-3 x^{3}+6}{x-3} $$

Graphs with Holes In this chapter we adopted the convention that in rational functions, the numerator and denominator don't share a common factor. In this exercise we consider the graph of a rational function that does not satisfy this rule. (a) Show that the graph of $$ r(x)=\frac{3 x^{2}-3 x-6}{x-2} $$ is the line \(y=3 x+3\) with the point \((2,9)\) removed. [Hint. Factor. What is the domain of \(r ? ]\) (b) Graph the rational functions: $$ \begin{aligned} s(x) &=\frac{x^{2}+x-20}{x+5} \\ t(x) &=\frac{2 x^{2}-x-1}{x-1} \\ u(x) &=\frac{x-2}{x^{2}-2 x} \end{aligned} $$

Find all rational zeros of the polynomial, and then find the irrational zeros, if any. Whenever appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes' Zule of Signs, the Uuadratic formula, or other factoring techniques. $$ P(x)=x^{5}-7 x^{4}+9 x^{3}+23 x^{2}-50 x+24 $$

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