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13- 30 . Factor the polynomial completely and find all its zeros. State the multiplicity of each zero. $$ Q(x)=x^{4}-1 $$

Short Answer

Expert verified
Zeros are 1, -1, i, -i, each with multiplicity 1.

Step by step solution

01

Recognize the Structure

The polynomial given is \( Q(x) = x^4 - 1 \). This is a difference of squares, as it can be seen in the form \( a^2 - b^2 \). In this case, \( a = x^2 \) and \( b = 1 \).
02

Apply Difference of Squares

The difference of squares formula is \( a^2 - b^2 = (a-b)(a+b) \). Applying this to \( x^4 - 1 \), we have \( (x^2 - 1)(x^2 + 1) \).
03

Factor Further, if Possible

Notice that \( x^2 - 1 \) can be further factored as another difference of squares: \( x^2 - 1 = (x-1)(x+1) \). However, \( x^2 + 1 \) cannot be factored further over the real numbers.
04

Write the Complete Factorization

Combining all factors, the complete factorization of the polynomial is \( Q(x) = (x-1)(x+1)(x^2 + 1) \).
05

Find the Zeros

Set each factor equal to zero to find the zeros. Solving \( x-1 = 0 \) gives \( x = 1 \), solving \( x+1 = 0 \) gives \( x = -1 \), and solving \( x^2 + 1 = 0 \) gives \( x^2 = -1 \), or \( x = i \) and \( x = -i \).
06

Determine Multiplicities

Each factor in the polynomial has degree 1, indicating that each zero \( x=1 \), \( x=-1 \), \( x=i \), and \( x=-i \) has a multiplicity of 1.

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Key Concepts

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

Difference of Squares
The **difference of squares** is a special algebraic pattern that simplifies certain expressions quickly. It is recognized in expressions of the form \(a^2 - b^2\). This pattern emerges because it can always be rewritten as the product of two binomials: \((a - b)(a + b)\).

In our example, the polynomial \(Q(x) = x^4 - 1\) fits this pattern because it can be seen as \((x^2)^2 - (1)^2\). Here, \(a = x^2\) and \(b = 1\). By applying the difference of squares, the polynomial becomes \((x^2 - 1)(x^2 + 1)\).

This method not only simplifies the expression but also helps in quickly identifying the roots by breaking down complex expressions into simpler parts. Note that while \(x^2 - 1\) can be further factored using the same pattern, \(x^2 + 1\) does not factor over the real numbers, indicating the presence of complex solutions.
Zeros of a Polynomial
The **zeros of a polynomial** are the values of \(x\) that make the polynomial equal to zero. These zeros can also be called roots or solutions.

To find these zeros, we set the expression equal to zero and solve. For \(Q(x) = (x-1)(x+1)(x^2 + 1)\), the zeros are found by setting each factor to zero:
  • \(x-1 = 0\) leads to \(x = 1\)
  • \(x+1 = 0\) leads to \(x = -1\)
  • \(x^2 + 1 = 0\) which simplifies to \(x^2 = -1\), resulting in \(x = i\) and \(x = -i\)

The roots \(x = 1\) and \(x = -1\) are real numbers, while \(x = i\) and \(x = -i\) are complex numbers, demonstrating that polynomials of higher degrees can have a mix of real and complex roots.
Multiplicity of Zeros
The **multiplicity of zeros** refers to how many times a particular solution occurs for a polynomial equation. It is related to the degree of the factor in the polynomial.

In the complete factorization \(Q(x) = (x-1)(x+1)(x^2 + 1)\), each factor \((x-1)\), \((x+1)\), and \((x^2 + 1)\) is only raised to the power of one. Therefore, each zero, \(x = 1\), \(x = -1\), \(x = i\), and \(x = -i\), has a multiplicity of 1.

This means each root is simple and only appears once in the solution set. Recognizing multiplicities is essential since it affects the graph's behavior near those roots - for an even multiplicity, the graph will touch and bounce off the x-axis, while for an odd multiplicity, it will cut through the x-axis.

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Most popular questions from this chapter

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. $$ P(x)=6 x^{4}-7 x^{3}-8 x^{2}+5 x $$

\(51-58=\) A polynomial \(P\) is given. (a) Find all the real zeros of \(P .\) (b) Sketch the graph of \(P\) . $$ P(x)=2 x^{3}-7 x^{2}+4 x+4 $$

How Many Real Zeros Can a Polynomial Have? Give examples of polynomials that have the following properties, or explain why it is impossible to find such a polynomial. (a) A polynomial of degree 3 that has no real zeros (b) A polynomial of degree 4 that has no real zeros (c) A polynomial of degree 3 that has three real zeros, only one of which is rational (d) A polynomial of degree 4 that has four real zeros, none of which is rational What must be true about the degree of a polynomial with integer coefficients if it has no real zeros?

The real solutions of the given equation are rational. List all possible rational roots using the Rational Zeros Theorem, and then graph the polynomial in the given viewing rectangle to determine which values are actually solutions. (All solutions can be seen in the given viewing rectangle.) $$ 2 x^{4}-5 x^{3}-14 x^{2}+5 x+12=0 ; \quad[-2,5] \text { by }[-40,40] $$

In Example 2 we saw that some simple rational functions can be graphed by shifting, stretching, or reflecting the graph of \(y=1 / x .\) In this exercise we consider rational functions that can be graphed by transforming the graph of \(y=1 / x^{2},\) shown on the following page. (a) Graph the function $$r(x)=\frac{1}{(x-2)^{2}}$$ by transforming the graph of \(y=1 / x^{2}\) (b) Use long division and factoring to show that the function $$s(x)=\frac{2 x^{2}+4 x+5}{x^{2}+2 x+1}$$ can be written as $$s(x)=2+\frac{3}{(x+1)^{2}}$$ Then graph \(s\) by transforming the graph of \(y=1 / x^{2}\) . (c) One of the following functions can be graphed by transforming the graph of \(y=1 / x^{2} ;\) the other cannot. Use transformations to graph the one that can be, and explain why this method doesn't work for the other one. $$p(x)=\frac{2-3 x^{2}}{x^{2}-4 x+4} \quad q(x)=\frac{12 x-3 x^{2}}{x^{2}-4 x+4}$$

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