3rd Edition

ISBN: 9781133947202

Author: Ron Larson

Publisher: Cengage Learning

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Question

Chapter 2.5, Problem 65E

To determine

To write : the polynomial as the product of linear factors and list all the zeros of function

Expert Solution & Answer

Answer to Problem 65E

The function as product of linear factors: $h(x)=(x−1−4i)(x−1+4i)$

The zeros of the function are: $x=1+4i ; 1−4i$

Explanation of Solution

Given information:

$h(x)=x2−2x+17$

Concept Involved:

Linear Factorization Theorem:If $f(x)$ is a polynomial of degree $n$ , where $n>0$ , then $f(x)$ has precisely $n$ linearfactors $f(x) = an(x − c1)(x − c2). . . (x − cn)$ where $c1, c2, . . . , cn$ are complex numbers.

Complex Zeros Occur in Conjugate Pairs: Let f be a polynomial function that has real coefficients. If $a+bi$ , where $b≠0$ , is a zero of the function, then the complex conjugates $a−bi$ is alsoa zero of the function.

Calculation:

To find the zeros of the function we need to set the function to zero $h(x)=0$

$x2−2x+17=0$

We can solve this equation using completing the square method by subtracting17 on both sides of the equation

$x2−2x+17−17=0−17$

Simplify on both sides of the equation

$x2−2x=−17$

In order to make the left side expression as perfect square trinomial, we need to add square of half of coefficient of x on both sides

$x2−2x+1=−17+1$

Simplify on right side of the equation

$x2−2x+1=−16$

Write the left side as a perfect square

$(x−1)2=−16$

Take square root on both sides

$(x−1)2=−16$

Split the number in the right side of the equation

$(x−1)2=−1⋅16$

Simplifying square root on both sides of the equation

$x−1 =−1⋅±4$

Replace $i$ for $−1$

$x−1=±4i$

Add 1 on both sides of the equation

$x−1+1=1±4i$

Simplify in left side of the equation

$x=1±4i$

List the zeros of the functions given:

$x=1+4i ; x=1−4i$

If $x=k$ is zero, then $x−k$ will be the factor of the function.

So $x−(1+4i)$ & $x−(1−4i)$ are the factors of the function

Write the function $h(x)=x2−2x+17$ as product of linear factors:

$h(x)=(x−1−4i)(x−1+4i)$

Conclusion:

The zeros of the given function $h(x)=x2−2x+17$ are: $x=1+4i ; 1−4i$

The function written as product of linear factors: $h(x)=(x−1−4i)(x−1+4i)$

Chapter 2.5, Problem 64E

Chapter 2.5, Problem 66E

Chapter 2 Solutions

Precalculus with Limits

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Prob. 73E Ch. 2.6 - Prob. 74E Ch. 2.6 - Prob. 75E Ch. 2.6 - Prob. 76E Ch. 2.6 - Prob. 77E Ch. 2.6 - Prob. 78E Ch. 2.6 - Prob. 79E Ch. 2.6 - Prob. 80E Ch. 2.6 - Prob. 81E Ch. 2.6 - Prob. 82E Ch. 2.7 - Prob. 1E Ch. 2.7 - Prob. 2E Ch. 2.7 - Prob. 3E Ch. 2.7 - Prob. 4E Ch. 2.7 - Prob. 5E Ch. 2.7 - Prob. 6E Ch. 2.7 - Prob. 7E Ch. 2.7 - Prob. 8E Ch. 2.7 - Prob. 9E Ch. 2.7 - Prob. 10E Ch. 2.7 - Prob. 11E Ch. 2.7 - Prob. 12E Ch. 2.7 - Prob. 13E Ch. 2.7 - Prob. 14E Ch. 2.7 - Prob. 15E Ch. 2.7 - Prob. 16E Ch. 2.7 - Prob. 17E Ch. 2.7 - Prob. 18E Ch. 2.7 - Prob. 19E Ch. 2.7 - Prob. 20E Ch. 2.7 - Prob. 21E Ch. 2.7 - Prob. 22E Ch. 2.7 - Prob. 23E Ch. 2.7 - Prob. 24E Ch. 2.7 - Prob. 25E Ch. 2.7 - Prob. 26E Ch. 2.7 - Prob. 27E Ch. 2.7 - Prob. 28E Ch. 2.7 - Prob. 29E Ch. 2.7 - Prob. 30E Ch. 2.7 - Prob. 31E Ch. 2.7 - Prob. 32E Ch. 2.7 - Prob. 33E Ch. 2.7 - Prob. 34E Ch. 2.7 - Prob. 35E Ch. 2.7 - Prob. 36E Ch. 2.7 - Prob. 37E Ch. 2.7 - Prob. 38E Ch. 2.7 - Prob. 39E Ch. 2.7 - Prob. 40E Ch. 2.7 - Prob. 41E Ch. 2.7 - Prob. 42E Ch. 2.7 - Prob. 43E Ch. 2.7 - Prob. 44E Ch. 2.7 - Prob. 45E Ch. 2.7 - Prob. 46E Ch. 2.7 - Prob. 47E Ch. 2.7 - Prob. 48E Ch. 2.7 - Prob. 49E Ch. 2.7 - Prob. 50E Ch. 2.7 - Prob. 51E Ch. 2.7 - Prob. 52E Ch. 2.7 - Prob. 53E Ch. 2.7 - Prob. 54E Ch. 2.7 - Prob. 55E Ch. 2.7 - Prob. 56E Ch. 2.7 - Prob. 57E Ch. 2.7 - Prob. 58E Ch. 2.7 - Prob. 59E Ch. 2.7 - Prob. 60E Ch. 2.7 - Prob. 61E Ch. 2.7 - Prob. 62E Ch. 2.7 - Prob. 63E Ch. 2.7 - Prob. 64E Ch. 2.7 - Prob. 65E Ch. 2.7 - Prob. 66E Ch. 2.7 - Prob. 67E Ch. 2.7 - Prob. 68E Ch. 2.7 - Prob. 69E Ch. 2.7 - Prob. 70E Ch. 2.7 - Prob. 71E Ch. 2.7 - Prob. 72E Ch. 2.7 - Prob. 73E Ch. 2.7 - Prob. 74E Ch. 2.7 - Prob. 75E Ch. 2.7 - Prob. 76E Ch. 2.7 - Prob. 77E Ch. 2.7 - Prob. 78E Ch. 2.7 - Prob. 79E Ch. 2.7 - Prob. 80E Ch. 2.7 - Prob. 81E Ch. 2.7 - Prob. 82E Ch. 2.7 - Prob. 83E Ch. 2.7 - Prob. 84E Ch. 2.7 - Prob. 85E Ch. 2.7 - Prob. 86E Ch. 2.7 - Prob. 87E Ch. 2.7 - Prob. 88E Ch. 2.7 - Prob. 89E Ch. 2.7 - Prob. 90E Ch. 2 - Prob. 1RE Ch. 2 - Prob. 2RE Ch. 2 - Prob. 3RE Ch. 2 - Prob. 4RE Ch. 2 - Prob. 5RE Ch. 2 - Prob. 6RE Ch. 2 - Prob. 7RE Ch. 2 - Prob. 8RE Ch. 2 - Prob. 9RE Ch. 2 - Prob. 10RE Ch. 2 - Prob. 11RE Ch. 2 - Prob. 12RE Ch. 2 - Prob. 13RE Ch. 2 - Prob. 14RE Ch. 2 - Prob. 15RE Ch. 2 - Prob. 16RE Ch. 2 - Prob. 17RE Ch. 2 - Prob. 18RE Ch. 2 - Prob. 19RE Ch. 2 - Prob. 20RE Ch. 2 - Prob. 21RE Ch. 2 - Prob. 22RE Ch. 2 - Prob. 23RE Ch. 2 - Prob. 24RE Ch. 2 - Prob. 25RE Ch. 2 - Prob. 26RE Ch. 2 - Prob. 27RE Ch. 2 - Prob. 28RE Ch. 2 - Prob. 29RE Ch. 2 - Prob. 30RE Ch. 2 - Prob. 31RE Ch. 2 - Prob. 32RE Ch. 2 - Prob. 33RE Ch. 2 - Prob. 34RE Ch. 2 - Prob. 35RE Ch. 2 - Prob. 36RE Ch. 2 - Prob. 37RE Ch. 2 - Prob. 38RE Ch. 2 - Prob. 39RE Ch. 2 - Prob. 40RE Ch. 2 - Prob. 41RE Ch. 2 - Prob. 42RE Ch. 2 - Prob. 43RE Ch. 2 - Prob. 44RE Ch. 2 - Prob. 45RE Ch. 2 - Prob. 46RE Ch. 2 - Prob. 47RE Ch. 2 - Prob. 48RE Ch. 2 - Prob. 49RE Ch. 2 - Prob. 50RE Ch. 2 - Prob. 51RE Ch. 2 - Prob. 52RE Ch. 2 - Prob. 53RE Ch. 2 - Prob. 54RE Ch. 2 - Prob. 55RE Ch. 2 - Prob. 56RE Ch. 2 - Prob. 57RE Ch. 2 - Prob. 58RE Ch. 2 - Prob. 59RE Ch. 2 - Prob. 60RE Ch. 2 - Prob. 61RE Ch. 2 - Prob. 62RE Ch. 2 - Prob. 63RE Ch. 2 - Prob. 64RE Ch. 2 - Prob. 65RE Ch. 2 - Prob. 66RE Ch. 2 - Prob. 1CT Ch. 2 - Prob. 2CT Ch. 2 - Prob. 3CT Ch. 2 - Prob. 4CT Ch. 2 - Prob. 5CT Ch. 2 - Prob. 6CT Ch. 2 - Prob. 7CT Ch. 2 - Prob. 8CT Ch. 2 - Prob. 9CT Ch. 2 - Prob. 10CT Ch. 2 - Prob. 11CT Ch. 2 - Prob. 12CT Ch. 2 - Prob. 13CT Ch. 2 - Prob. 14CT Ch. 2 - Prob. 15CT Ch. 2 - Prob. 16CT Ch. 2 - Prob. 17CT Ch. 2 - Prob. 18CT Ch. 2 - Prob. 1PS Ch. 2 - Prob. 2PS Ch. 2 - Prob. 3PS Ch. 2 - Prob. 4PS Ch. 2 - Prob. 5PS Ch. 2 - Prob. 6PS Ch. 2 - Prob. 7PS Ch. 2 - Prob. 8PS Ch. 2 - Prob. 9PS Ch. 2 - Prob. 10PS Ch. 2 - Prob. 11PS Ch. 2 - Prob. 12PS Ch. 2 - Prob. 13PS

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