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Factor the polynomial and use the factored form to find the zeros. Then sketch the graph. $$P(x)=x^{4}-2 x^{3}+8 x-16$$

Short Answer

Expert verified
The zeros are \(x = 2\) and \(x = -2\). The complete factored form is \((x - 2)(x + 2)(x^2 - 2x + 4)\).

Step by step solution

01

Look for Common Factors

First, examine the polynomial \( P(x) = x^{4} - 2x^{3} + 8x - 16 \) to see if there are any common factors across all terms. There are no common factors in this expression.
02

Attempt Polynomial Factoring by Grouping

To factor by grouping, divide the polynomial into two pairs: \((x^4 - 2x^3)\) and \((8x - 16)\). The first group can be factored as \((x^3)(x - 2)\) and the second group as \(8(x - 2)\).
03

Factor Out the Common Binomial

Notice that both terms \(x^3(x - 2)\) and \(8(x - 2)\) share a common binomial factor \((x - 2)\). Factor \((x - 2)\) out, resulting in \((x - 2)(x^3 + 8)\).
04

Further Factor the Remaining Cubic Term

The term \(x^3 + 8\) is a sum of cubes. Use the sum of cubes formula \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\), where \(a = x\) and \(b = 2\). This gives \( (x + 2)(x^2 - 2x + 4) \).
05

Write the Complete Factored Form

Combine the factored components from the previous steps: \((x - 2)(x + 2)(x^2 - 2x + 4)\).
06

Find the Zeros of the Polynomial

Set each factor equal to zero: \(x - 2 = 0\) gives \(x = 2\), \(x + 2 = 0\) gives \(x = -2\). The quadratic \(x^2 - 2x + 4\) does not have real roots, as its discriminant \((-2)^2 - 4 \times 1 \times 4 = -12\) is negative.
07

Sketch the Graph

Plot the polynomial, marking its zeros at \(x = 2\) and \(x = -2\). The graph has turning points or inflection points but does not cross the x-axis at any other points due to the complex roots of the remaining quadratic.

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Key Concepts

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

Zeros of Polynomials
When dealing with polynomials, identifying the zeros is a key step. Zeros are the values of the variable that make the polynomial equal to zero. They are also known as the roots of the polynomial. For example, if given a polynomial like \( P(x) = x^{4} - 2x^{3} + 8x - 16 \), the goal is to express it in its factored form and identify values of \( x \) that satisfy the equation \( P(x) = 0 \).

The factorization process might involve steps like factoring by grouping or using identities. After factoring, each binomial can be set to zero individually. For the polynomial \((x - 2)(x + 2)(x^2 - 2x + 4)\), setting each factor to zero yields the zeros. Here, \( x = 2 \) and \( x = -2 \) are real zeros, and the expression \( x^2 - 2x + 4 \) does not factor further into real roots, as it has a negative discriminant.
Sum of Cubes
The sum of cubes is a special factorization form used to handle expressions like \( x^3 + 8 \). Recognizing this pattern allows us to utilize a specific formula, \( a^3 + b^3 = (a+b)(a^2-ab+b^2) \).

Here, \( a \) is \( x \) and \( b \) is \( 2 \), transforming the expression into \((x + 2)(x^2 - 2x + 4)\). This approach simplifies the factorization of complex cubic terms, lending us a way to break down the polynomial into simpler factors for easier handling, especially when searching for zeros.
Factoring by Grouping
Factoring by grouping is often a useful technique when dealing with polynomials that do not show any obvious common factors. It’s a strategy that works well when the polynomial can be divided into pairs or groups that have a common factor.

For the polynomial \( P(x) = x^{4} - 2x^{3} + 8x - 16 \), we can separate it into groups: \((x^4 - 2x^3)\) and \((8x - 16)\). Within these subgroups, common factors \( x^3 \) and \( 8 \), respectively, can be extracted, giving us \( x^3(x - 2) + 8(x - 2) \). Notice that \( (x - 2) \) is a common binomial.

Thus, we can factor it out further as \((x - 2)(x^3 + 8)\), paving the way for more detailed factoring operations, such as the sum of cubes.
Graphing Polynomials
Graphing polynomials involves plotting the curve represented by the polynomial equation. The zeros of the polynomial give critical points where the graph intersects the x-axis. These are the points \( (2, 0) \) and \( (-2, 0) \) for this polynomial.

Since this polynomial is of degree four, we can expect the graph to have up to four intercepts and several turns. However, not all factors provide real roots, as shown by \( x^2 - 2x + 4 \), which remains above the x-axis because of its complex roots.
  • The graph crosses the x-axis at \( x = 2 \) and \( x = -2 \).
  • It will have turning points which we can find by examining the behavior of the function near the zeros.
  • The polynomial's shape (U-curve, inverted U, etc.) depends on the leading term, \( x^4 \), indicating the end behavior will extend towards positive infinity at both ends.
This analysis helps sketch an approximate graph representing how the polynomial behaves across different intervals.

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Most popular questions from this chapter

Find the slant asymptote, the vertical asymptotes, and sketch a graph of the function. $$r(x)=\frac{x^{2}+2 x}{x-1}$$

Give an example of a rational function that has vertical asymptote \(x=3 .\) Now give an example of one that has vertical asymptote \(x=3\) and horizontal asymptote \(y=2\) Now give an example of a rational function with vertical asymptotes \(x=1\) and \(x=-1,\) horizontal asymptote \(y=0,\) and \(x\) -intercept 4.

Graph the rational function, and find all vertical asymptotes, \(x\) - and \(y\) -intercepts, and local extrema, comect 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. $$r(x)=\frac{4+x^{2}-x^{4}}{x^{2}-1}$$

Find all solutions of the equation and express them in the form \(a+b i\) $$t+3+\frac{3}{t}=0$$

The quadratic formula can be used to solve any quadratic (or second-degree) equation. You might have wondered whether similar formulas exist for cubic (thirddegree), quartic (fourth-degree), and higher-degree equations. For the depressed cubic \(x^{3}+p x+q=0,\) Cardano (page 274) found the following formula for one solution: $$x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$$ A formula for quartic equations was discovered by the Italian mathematician Ferrari in \(1540 .\) In 1824 the Norwegian mathematician Niels Henrik Abel proved that it is impossible to write a quintic formula, that is, a formula for fifth-degree equations. Finally, Galois (page 254) gave a criterion for determining which equations can be solved by a formula involving radicals. Use the cubic formula to find a solution for the following equations. Then solve the equations using the methods you learned in this section. Which method is easier? (a) \(x^{3}-3 x+2=0\) (b) \(x^{3}-27 x-54=0\) (c) \(x^{3}+3 x+4=0\)

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