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Chapter 3: Problem 18

Factor the polynomial completely and find all its zeros. State the multiplicity of each zero. $$P(x)=x^{3}-x^{2}+x$$

Short Answer

Expert verified
The zeros are 0 (multiplicity 1), \(\frac{1+i\sqrt{3}}{2}\) (multiplicity 1), and \(\frac{1-i\sqrt{3}}{2}\) (multiplicity 1).

Step by step solution

01

Factor Out the Greatest Common Factor (GCF)

The given polynomial is \( P(x) = x^3 - x^2 + x \). First, identify the greatest common factor present in all terms. Here, each term contains at least one \( x \), so we can factor out an \( x \):\[ P(x) = x(x^2 - x + 1) \].
02

Attempt Further Factoring

Now, look at the quadratic expression \( x^2 - x + 1 \) to see if it can be factored further. The discriminant method can be used: calculate \( b^2 - 4ac \) where \( a=1, b=-1, c=1 \). Calculate \((-1)^2 - 4 \times 1 \times 1 = 1 - 4 = -3\), which is less than zero, indicating that \( x^2 - x + 1 \) does not factor over the real numbers.
03

Find the Zeros of the Polynomial

The polynomial can be split into the factors we have found: \( x(x^2-x+1) \). The zero of the first factor \( x \) is obvious: \( x=0 \). For the quadratic \( x^2-x+1 \), since it does not factor further, find its zeros using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
04

Solve for the Non-Real Zeros

Using the quadratic formula with \( a=1, b=-1, c=1 \), we plug in the values: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2-4\cdot1\cdot1}}{2\cdot1} = \frac{1 \pm \sqrt{-3}}{2}\], which simplifies to \[ x = \frac{1 \pm i\sqrt{3}}{2} \]. These are the complex zeros.
05

State the Zeros and Their Multiplicities

Since \( x=0 \) comes from a linear factor \( x \), it has multiplicity 1. The complex solutions \( \frac{1+i\sqrt{3}}{2} \) and \( \frac{1-i\sqrt{3}}{2} \) also have multiplicities of 1, based on the quadratic factor.

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Key Concepts

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

Zeros of Polynomials
Understanding the zeros of a polynomial is key to grasping how a polynomial behaves and intersects the x-axis. In simpler terms, zeros are the values of \( x \) that make the polynomial equal to zero. These are often where graphs of polynomials cross or touch the x-axis.
  • To find zeros, we set each factor of the polynomial to zero and solve for \( x \).
  • If a polynomial can be factored, like \( P(x) = x(x^2 - x + 1) \), we identify its zeros from the factors. The factor \( x \) gives a zero at \( x = 0 \).
  • Any polynomial, say of degree \( n \), has at most \( n \) real zeros. However, complex numbers might be involved, especially when the discriminant is negative, as seen in this example.
Recognizing zeros helps you understand the shape and crosses of the polynomial graph, and they tell a story about the factors in the expression.
Quadratic Formula
The quadratic formula is your go-to tool when you're unable to factor a quadratic expression easily. This magical formula allows you to find zeros of quadratic equations in the form \( ax^2 + bx + c \). It's given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Let's dissect it:
  • The "\( -b \pm \)" part accounts for both potential solutions of the quadratic. Some quadratics might only have one solution when the discriminant is zero, as both solutions merge into one.
  • The value inside the square root, \( b^2 - 4ac \), known as the discriminant, decides the nature of the roots.
  • Positive discriminant means two real solutions, zero means one real solution, and negative means two complex solutions.
In our example \( x^2 - x + 1 \), the discriminant was negative (\(-3\)), leading to complex zeros. We calculated these using the formula, providing imaginary parts that indicate the zeros don't intersect the x-axis on a real plane.
Multiplicity of Zeros
The multiplicity of zeros indicates how many times a particular zero appears as a solution for the polynomial. This concept also describes how the graph of the function interacts with the x-axis at that point.
  • If a zero has a multiplicity of 1, the graph crosses the x-axis at that point. This is what occurs with \( x = 0 \) for our polynomial \( P(x) \).
  • Higher multiplicities, such as 2 or more, mean the zero is repeated, and the graph might just touch the x-axis and turn back. For instance, a multiplicity of 2 implies the graph "bounces" off the axis.
  • Multiplicity can have significant implications. It affects the shape and symmetry of the polynomial graph. Multiplicity gives insight into the "behavior" of how a function ascends or descends relative to the axis.
Knowing the multiplicities helps further unravel the full "geometry" of polynomials鈥攊mproving understanding of graph turning points and symmetry.

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Most popular questions from this chapter

Find integers that are upper and lower bounds for the real zeros of the polynomial. $$P(x)=x^{5}-x^{4}+1$$

Graph the rational function \(f\) and determine all vertical asymptotes from your graph. Then graph \(f\) and \(g\) in a sufficiently large viewing rectangle to show that they have the same end behavior. $$f(x)=\frac{-x^{3}+6 x^{2}-5}{x^{2}-2 x}, g(x)=-x+4$$

Find the slant asymptote, the vertical asymptotes, and sketch a graph of the function. $$r(x)=\frac{x^{3}+x^{2}}{x^{2}-4}$$

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' Rule of Signs, the quadratic formula, or other factoring techniques. $$P(x)=8 x^{5}-14 x^{4}-22 x^{3}+57 x^{2}-35 x+6$$

Find all horizontal and vertical asymptotes (if any). $$r(x)=\frac{x^{3}+3 x^{2}}{x^{2}-4}$$
See all solutions

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