EBK PRECALCULUS W/LIMITS
EBK PRECALCULUS W/LIMITS
4th Edition
ISBN: 9781337516853
Author: Larson
Publisher: CENGAGE CO
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Chapter 2.5, Problem 22E
To determine

Find the possible rational zeros of the function, use the Rational Zero Test.

Expert Solution & Answer
Check Mark

Answer to Problem 22E

The rational zeros of the function are 2,5,3 .

Explanation of Solution

Given information:

The functionis h(x)=x319x+30 .

Formula Used:

Rational Zero Test.

The formula is: Possiblerationalzeros=FactorsofconstanttermFactorsofleadingcoefficient .

To find the zeros of the function, substitute the function f(x) equal to 0 and solve it.

Calculation:

The function is h(x)=x319x+30 .  ......(1)

To find the zeros of function, use the Rational Zero Test.

The leading coefficient is 1 and the constant term is 30 .

  Possiblerationalzeros=,Factorsof30Factorsof1=±1,±2,±3,±5,±6,±10,±15,±30±1=±1,±2,±3,±5,±6,±10,±15,±30

Put x=2 in the function (1).

  h(2)=(2)319(2)+30=838+30=0

Thus, x=2 is a zero of the function h(x) .

To simplify the function, take x=2 as a rational root and use synthetic division rule.

  2|1019302430_12150

Rewrite the above polynomial as:

  h(x)=(x2)(x2+2x15)

To solve the above quadratic equation, split the middle term.

  h(x)=(x2)(x2+5x3x15)=(x2)(x(x+5)3(x+5))=(x2)(x+5)(x3)

To find the zeros of the function, substitute the function h(x) equal to 0 and solve it.

  (x2)(x+5)(x3)=0

Rewrite it as:

  (x2)=0or(x+5)=0or(x3)=0

That is,

  (x2)=0x=2

Or

  (x+5)=0x=5

Or

  (x3)=0x=3

Hence, the rational zeros of the function are 2,5,3 .

Chapter 2 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 2.1 - Prob. 11ECh. 2.1 - Prob. 12ECh. 2.1 - Prob. 13ECh. 2.1 - Prob. 14ECh. 2.1 - Prob. 15ECh. 2.1 - Prob. 16ECh. 2.1 - Prob. 17ECh. 2.1 - Prob. 18ECh. 2.1 - Prob. 19ECh. 2.1 - Prob. 20ECh. 2.1 - Prob. 21ECh. 2.1 - Prob. 22ECh. 2.1 - Prob. 23ECh. 2.1 - Prob. 24ECh. 2.1 - Prob. 25ECh. 2.1 - Prob. 26ECh. 2.1 - Prob. 27ECh. 2.1 - Prob. 28ECh. 2.1 - Prob. 29ECh. 2.1 - Prob. 30ECh. 2.1 - Prob. 31ECh. 2.1 - Prob. 32ECh. 2.1 - Prob. 33ECh. 2.1 - Prob. 34ECh. 2.1 - Prob. 35ECh. 2.1 - Prob. 36ECh. 2.1 - Prob. 37ECh. 2.1 - Prob. 38ECh. 2.1 - Prob. 39ECh. 2.1 - Prob. 40ECh. 2.1 - Prob. 41ECh. 2.1 - Prob. 42ECh. 2.1 - Prob. 43ECh. 2.1 - Prob. 44ECh. 2.1 - Prob. 45ECh. 2.1 - Prob. 46ECh. 2.1 - Prob. 47ECh. 2.1 - Prob. 48ECh. 2.1 - Prob. 49ECh. 2.1 - Prob. 50ECh. 2.1 - Prob. 51ECh. 2.1 - Prob. 52ECh. 2.1 - Prob. 53ECh. 2.1 - Prob. 54ECh. 2.1 - Prob. 55ECh. 2.1 - Prob. 56ECh. 2.1 - Prob. 57ECh. 2.1 - Prob. 58ECh. 2.1 - Prob. 59ECh. 2.1 - Prob. 60ECh. 2.1 - Prob. 61ECh. 2.1 - Prob. 62ECh. 2.1 - Prob. 63ECh. 2.1 - 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Prob. 6ECh. 2.5 - Prob. 7ECh. 2.5 - Prob. 8ECh. 2.5 - Prob. 9ECh. 2.5 - Prob. 10ECh. 2.5 - Prob. 11ECh. 2.5 - Prob. 12ECh. 2.5 - Prob. 13ECh. 2.5 - Prob. 14ECh. 2.5 - Prob. 15ECh. 2.5 - Prob. 16ECh. 2.5 - Prob. 17ECh. 2.5 - Prob. 18ECh. 2.5 - Prob. 19ECh. 2.5 - Prob. 20ECh. 2.5 - Prob. 21ECh. 2.5 - Prob. 22ECh. 2.5 - Prob. 23ECh. 2.5 - Prob. 24ECh. 2.5 - Prob. 25ECh. 2.5 - Prob. 26ECh. 2.5 - Prob. 27ECh. 2.5 - Prob. 28ECh. 2.5 - Prob. 29ECh. 2.5 - Prob. 30ECh. 2.5 - Prob. 31ECh. 2.5 - Prob. 32ECh. 2.5 - Prob. 33ECh. 2.5 - Prob. 34ECh. 2.5 - Prob. 35ECh. 2.5 - Prob. 36ECh. 2.5 - Prob. 37ECh. 2.5 - Prob. 38ECh. 2.5 - Prob. 39ECh. 2.5 - Prob. 40ECh. 2.5 - Prob. 41ECh. 2.5 - Prob. 42ECh. 2.5 - Prob. 43ECh. 2.5 - Prob. 44ECh. 2.5 - Prob. 45ECh. 2.5 - Prob. 46ECh. 2.5 - Prob. 47ECh. 2.5 - Prob. 48ECh. 2.5 - Prob. 49ECh. 2.5 - Prob. 50ECh. 2.5 - Prob. 51ECh. 2.5 - Prob. 52ECh. 2.5 - Prob. 53ECh. 2.5 - Prob. 54ECh. 2.5 - Prob. 55ECh. 2.5 - Prob. 56ECh. 2.5 - Prob. 57ECh. 2.5 - Prob. 58ECh. 2.5 - 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10 - Roots of polynomials; Author: Technion;https://www.youtube.com/watch?v=88YUeigknNg;License: Standard YouTube License, CC-BY