PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Chapter 2, Problem 98CR

(a)

To determine

To find : All the zeros of function f(x)=−3x3−19x2−4x+12

(a)

Expert Solution
Check Mark

Answer to Problem 98CR

  x=−1, x=23, x=−6

Explanation of Solution

Given information :

  f(x)=−3x3−19x2−4x+12

Concept used:

Zeros of polynomial: -

It is defined by as any real value of x , for which the value of polynomial becomes zero.

Rational root theorem:

If the polynomial P(x)=anxn+an−1xn−1+...+a2x2+a1x+a0 has any rational roots, then they must be of the form ±{factor of a0factor of an}

Calculation:

  f(x)=−3x3−19x2−4x+12f(x)=−(3x3+19x2+4x−12)

Since, all coefficients are integers.

So, apply rational zeros theorem

The constant term is 51 and factor of 51 is ±1, ±2, ±3, ±4,±6, ±12 . These are the possible values for f(x)

If a is a root of the polynomial f(x) , then the remainder from the division of f(x) by x−a should be equal 0

Check:

  −1 : divide f(x) by x+1

  x+13x2+16x−123x3+19x2+4x−123x3+3x2−     −_       16x2+4x−12       16x2+16x−         −_                   −12x+12                   −12x+12+         −_                               0     

The quotient is 3x2+16x−12 , and the remainder is 0

Since the remainder is 0 , then −1 is the root, and x+1 is the factor

  −(3x3+19x2+4x−12)=−(x+1)(3x2+16x−12)

Now, find the roots of 3x2+16x−12

  For a quadratic equation of the form ax2+bx+c=0 the solutions are x1, 2=−b±b2−4ac2aFor  a=3, b=16, c=−12: x1, 2=−16±162−4⋅ 3(−12)2⋅ 3x1, 2=−16±256+1446x1, 2=−16±4006x1, 2=−16+206,−16−206x=23, x=−6

So, the zeros of function f(x) is

  x=−1, x=23, x=−6

(b)

To determine

To find : linear factor of f(x)=−3x3−19x2−4x+12

(b)

Expert Solution
Check Mark

Answer to Problem 98CR

  −(x+1)(x+6)(3x−2)

Explanation of Solution

Given information :

  f(x)=−3x3−19x2−4x+12

Concept used:

Zeros of polynomial: -

It is defined by as any real value of x , for which the value of polynomial becomes zero.

Rational root theorem:

If the polynomial P(x)=anxn+an−1xn−1+...+a2x2+a1x+a0 has any rational roots, then they must be of the form ±{factor of a0factor of an}

Calculation:

  f(x)=−3x3−19x2−4x+12f(x)=−(3x3+19x2+4x−12)

Since, all coefficients are integers.

So, apply rational zeros theorem

The constant term is 51 and factor of 51 is ±1, ±2, ±3, ±4,±6, ±12 . These are the possible values for f(x)

If a is a root of the polynomial f(x) , then the remainder from the division of f(x) by x−a should be equal 0

Check −1 : divide f(x) by x+1

  x+13x2+16x−123x3+19x2+4x−123x3+3x2−     −_       16x2+4x−12       16x2+16x−         −_                   −12x+12                   −12x+12+         −_                               0     

The quotient is 3x2+16x−12 , and the remainder is 0

Since the remainder is 0 , then −1 is the root, and x+1 is the factor

  −(3x3+19x2+4x−12)=−(x+1)(3x2+16x−12)

Now, factor 3x2+16x−12

  3x2+16x−12=3x2+18x−2x−12                              =3x(x+6)−2(x+6)                              =(x+6)(3x−2)

So, the linear factorization of f(x) is

  −(x+1)(x+6)(3x−2)

(c)

To determine

To find : x-intercepts using the factorization of f(x)=−3x3−19x2−4x+12 and use graphing utility to verify that the real zeros are the only x-intercepts

(c)

Expert Solution
Check Mark

Answer to Problem 98CR

  x=(−1,0),  (−6,0) and (23,0)

Explanation of Solution

Given information :

  f(x)=−3x3−19x2−4x+12

Concept used:

Zeros of polynomial: -

It is defined by as any real value of x , for which the value of polynomial becomes zero.

Rational root theorem:

If the polynomial P(x)=anxn+an−1xn−1+...+a2x2+a1x+a0 has any rational roots, then they must be of the form ±{factor of a0factor of an}

Calculation:

  f(x)=−3x3−19x2−4x+12f(x)=−(3x3+19x2+4x−12)

Since, all coefficients are integers.

So, apply rational zeros theorem

The constant term is 51 and factor of 51 is ±1, ±2, ±3, ±4,±6, ±12 . These are the possible values for f(x)

If a is a root of the polynomial f(x) , then the remainder from the division of f(x) by x−a should be equal 0

Check −1 : divide f(x) by x+1

  x+13x2+16x−123x3+19x2+4x−123x3+3x2−     −_       16x2+4x−12       16x2+16x−         −_                   −12x+12                   −12x+12+         −_                               0     

The quotient is 3x2+16x−12 , and the remainder is 0

Since the remainder is 0 , then −1 is the root, and x+1 is the factor

  −(3x3+19x2+4x−12)=−(x+1)(3x2+16x−12)

Now, factor 3x2+16x−12

  3x2+16x−12=3x2+18x−2x−12                              =3x(x+6)−2(x+6)                              =(x+6)(3x−2)

So, the linear factorization of f(x) is

  −(x+1)(x+6)(3x−2)=0x+1=0⇒x=−1x+6=0⇒x=−63x−2=0⇒x=23

Graph:

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 2, Problem 98CR

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PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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Prob. 98ECh. 2.5 - Prob. 1ECh. 2.5 - Prob. 2ECh. 2.5 - Prob. 3ECh. 2.5 - Prob. 4ECh. 2.5 - Prob. 5ECh. 2.5 - 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 - Prob. 59ECh. 2.5 - Prob. 60ECh. 2.5 - Prob. 61ECh. 2.5 - Prob. 62ECh. 2.5 - Prob. 63ECh. 2.5 - Prob. 64ECh. 2.5 - Prob. 65ECh. 2.5 - Prob. 66ECh. 2.5 - Prob. 67ECh. 2.5 - Prob. 68ECh. 2.5 - Prob. 69ECh. 2.5 - Prob. 70ECh. 2.5 - Prob. 71ECh. 2.5 - Prob. 72ECh. 2.5 - Prob. 73ECh. 2.5 - Prob. 74ECh. 2.5 - Prob. 75ECh. 2.5 - Prob. 76ECh. 2.5 - Prob. 77ECh. 2.5 - Prob. 78ECh. 2.6 - Prob. 1ECh. 2.6 - Prob. 2ECh. 2.6 - Prob. 3ECh. 2.6 - Prob. 4ECh. 2.6 - Prob. 5ECh. 2.6 - Prob. 6ECh. 2.6 - Prob. 7ECh. 2.6 - Prob. 8ECh. 2.6 - Prob. 9ECh. 2.6 - Prob. 10ECh. 2.6 - Prob. 11ECh. 2.6 - Prob. 12ECh. 2.6 - Prob. 13ECh. 2.6 - Prob. 14ECh. 2.6 - Prob. 15ECh. 2.6 - Prob. 16ECh. 2.6 - Prob. 17ECh. 2.6 - Prob. 18ECh. 2.6 - Prob. 19ECh. 2.6 - Prob. 20ECh. 2.6 - Prob. 21ECh. 2.6 - Prob. 22ECh. 2.6 - Prob. 23ECh. 2.6 - Prob. 24ECh. 2.6 - Prob. 25ECh. 2.6 - Prob. 26ECh. 2.6 - Prob. 27ECh. 2.6 - Prob. 28ECh. 2.6 - Prob. 29ECh. 2.6 - Prob. 30ECh. 2.6 - Prob. 31ECh. 2.6 - Prob. 32ECh. 2.6 - Prob. 33ECh. 2.6 - Prob. 34ECh. 2.6 - Prob. 35ECh. 2.6 - Prob. 36ECh. 2.6 - Prob. 37ECh. 2.6 - Prob. 38ECh. 2.6 - Prob. 39ECh. 2.6 - Prob. 40ECh. 2.6 - Prob. 41ECh. 2.6 - Prob. 42ECh. 2.6 - Prob. 43ECh. 2.6 - Prob. 44ECh. 2.6 - Prob. 45ECh. 2.6 - Prob. 46ECh. 2.6 - Prob. 47ECh. 2.6 - Prob. 48ECh. 2.6 - Prob. 49ECh. 2.6 - Prob. 50ECh. 2.6 - Prob. 51ECh. 2.6 - Prob. 52ECh. 2.6 - Prob. 53ECh. 2.6 - Prob. 54ECh. 2.6 - Prob. 55ECh. 2.6 - Prob. 56ECh. 2.6 - Prob. 57ECh. 2.6 - Prob. 58ECh. 2.6 - Prob. 59ECh. 2.6 - Prob. 60ECh. 2.6 - Prob. 61ECh. 2.6 - Prob. 62ECh. 2.6 - Prob. 63ECh. 2.6 - Prob. 64ECh. 2.6 - Prob. 65ECh. 2.6 - Prob. 66ECh. 2.7 - Prob. 1ECh. 2.7 - Prob. 2ECh. 2.7 - Prob. 3ECh. 2.7 - Prob. 4ECh. 2.7 - Prob. 5ECh. 2.7 - Prob. 6ECh. 2.7 - Prob. 7ECh. 2.7 - Prob. 8ECh. 2.7 - Prob. 9ECh. 2.7 - Prob. 10ECh. 2.7 - Prob. 11ECh. 2.7 - Prob. 12ECh. 2.7 - Prob. 13ECh. 2.7 - Prob. 14ECh. 2.7 - Prob. 15ECh. 2.7 - Prob. 16ECh. 2.7 - Prob. 17ECh. 2.7 - Prob. 18ECh. 2.7 - Prob. 19ECh. 2.7 - Prob. 20ECh. 2.7 - Prob. 21ECh. 2.7 - Prob. 22ECh. 2.7 - Prob. 23ECh. 2.7 - Prob. 24ECh. 2.7 - Prob. 25ECh. 2.7 - Prob. 26ECh. 2.7 - Prob. 27ECh. 2.7 - Prob. 28ECh. 2.7 - Prob. 29ECh. 2.7 - Prob. 30ECh. 2.7 - Prob. 31ECh. 2.7 - Prob. 32ECh. 2.7 - Prob. 33ECh. 2.7 - Prob. 34ECh. 2.7 - Prob. 35ECh. 2.7 - Prob. 36ECh. 2.7 - Prob. 37ECh. 2.7 - Prob. 38ECh. 2.7 - Prob. 39ECh. 2.7 - Prob. 40ECh. 2.7 - Prob. 41ECh. 2.7 - Prob. 42ECh. 2.7 - Prob. 43ECh. 2.7 - Prob. 44ECh. 2.7 - Prob. 45ECh. 2.7 - Prob. 46ECh. 2.7 - Prob. 47ECh. 2.7 - Prob. 48ECh. 2.7 - Prob. 49ECh. 2.7 - Prob. 50ECh. 2.7 - Prob. 51ECh. 2.7 - Prob. 52ECh. 2.7 - Prob. 53ECh. 2.7 - Prob. 54ECh. 2.7 - Prob. 55ECh. 2.7 - Prob. 56ECh. 2.7 - Prob. 57ECh. 2.7 - Prob. 58ECh. 2.7 - Prob. 59ECh. 2.7 - Prob. 60ECh. 2.7 - Prob. 61ECh. 2.7 - Prob. 62ECh. 2.7 - Prob. 63ECh. 2.7 - Prob. 64ECh. 2.7 - Prob. 65ECh. 2.7 - Prob. 66ECh. 2.7 - Prob. 67ECh. 2.7 - Prob. 68ECh. 2.7 - Prob. 69ECh. 2.7 - Prob. 70ECh. 2.7 - Prob. 71ECh. 2.7 - Prob. 72ECh. 2.7 - Prob. 73ECh. 2.7 - Prob. 74ECh. 2.7 - Prob. 75ECh. 2.7 - Prob. 76ECh. 2.7 - Prob. 77ECh. 2.7 - Prob. 78ECh. 2.7 - Prob. 79ECh. 2.7 - Prob. 80ECh. 2.7 - Prob. 81ECh. 2.7 - Prob. 82ECh. 2.7 - Prob. 83ECh. 2.7 - Prob. 84ECh. 2.7 - Prob. 85ECh. 2.7 - Prob. 86ECh. 2.7 - Prob. 87ECh. 2.7 - Prob. 88ECh. 2.7 - Prob. 89ECh. 2.7 - Prob. 90ECh. 2.7 - Prob. 91ECh. 2.7 - Prob. 92ECh. 2.7 - Prob. 93ECh. 2.7 - Prob. 94ECh. 2.7 - Prob. 95ECh. 2.7 - Prob. 96ECh. 2.7 - Prob. 97ECh. 2.7 - Prob. 98ECh. 2.7 - Prob. 99ECh. 2.7 - Prob. 100ECh. 2.7 - Prob. 101ECh. 2.8 - Prob. 1ECh. 2.8 - Prob. 2ECh. 2.8 - Prob. 3ECh. 2.8 - Prob. 4ECh. 2.8 - Prob. 5ECh. 2.8 - Prob. 6ECh. 2.8 - Prob. 7ECh. 2.8 - Prob. 8ECh. 2.8 - Prob. 9ECh. 2.8 - Prob. 10ECh. 2.8 - Prob. 11ECh. 2.8 - Prob. 12ECh. 2.8 - Prob. 13ECh. 2.8 - Prob. 14ECh. 2.8 - Prob. 15ECh. 2.8 - Prob. 16ECh. 2.8 - Prob. 17ECh. 2.8 - Prob. 18ECh. 2.8 - Prob. 19ECh. 2.8 - Prob. 20ECh. 2.8 - Prob. 21ECh. 2.8 - Prob. 22ECh. 2.8 - Prob. 23ECh. 2.8 - Prob. 24ECh. 2.8 - Prob. 25ECh. 2.8 - Prob. 26ECh. 2.8 - Prob. 27ECh. 2.8 - Prob. 28ECh. 2.8 - Prob. 29ECh. 2.8 - Prob. 30ECh. 2.8 - Prob. 31ECh. 2.8 - Prob. 32ECh. 2.8 - Prob. 33ECh. 2.8 - Prob. 34ECh. 2.8 - Prob. 35ECh. 2.8 - Prob. 36ECh. 2.8 - Prob. 37ECh. 2.8 - Prob. 38ECh. 2.8 - Prob. 39ECh. 2.8 - Prob. 40ECh. 2 - Prob. 1CRCh. 2 - Prob. 2CRCh. 2 - Prob. 3CRCh. 2 - Prob. 4CRCh. 2 - Prob. 5CRCh. 2 - Prob. 6CRCh. 2 - Prob. 7CRCh. 2 - Prob. 8CRCh. 2 - Prob. 9CRCh. 2 - Prob. 10CRCh. 2 - Prob. 11CRCh. 2 - Prob. 12CRCh. 2 - Prob. 13CRCh. 2 - Prob. 14CRCh. 2 - Prob. 15CRCh. 2 - Prob. 16CRCh. 2 - Prob. 17CRCh. 2 - Prob. 18CRCh. 2 - Prob. 19CRCh. 2 - Prob. 20CRCh. 2 - Prob. 21CRCh. 2 - Prob. 22CRCh. 2 - Prob. 23CRCh. 2 - Prob. 24CRCh. 2 - Prob. 25CRCh. 2 - Prob. 26CRCh. 2 - Prob. 27CRCh. 2 - Prob. 28CRCh. 2 - Prob. 29CRCh. 2 - Prob. 30CRCh. 2 - Prob. 31CRCh. 2 - Prob. 32CRCh. 2 - Prob. 33CRCh. 2 - Prob. 34CRCh. 2 - Prob. 35CRCh. 2 - Prob. 36CRCh. 2 - Prob. 37CRCh. 2 - Prob. 38CRCh. 2 - Prob. 39CRCh. 2 - Prob. 40CRCh. 2 - Prob. 41CRCh. 2 - Prob. 42CRCh. 2 - Prob. 43CRCh. 2 - Prob. 44CRCh. 2 - Prob. 45CRCh. 2 - Prob. 46CRCh. 2 - Prob. 47CRCh. 2 - Prob. 48CRCh. 2 - Prob. 49CRCh. 2 - Prob. 50CRCh. 2 - Prob. 51CRCh. 2 - Prob. 52CRCh. 2 - Prob. 53CRCh. 2 - Prob. 54CRCh. 2 - Prob. 55CRCh. 2 - Prob. 56CRCh. 2 - Prob. 57CRCh. 2 - Prob. 58CRCh. 2 - Prob. 59CRCh. 2 - Prob. 60CRCh. 2 - Prob. 61CRCh. 2 - Prob. 62CRCh. 2 - Prob. 63CRCh. 2 - Prob. 64CRCh. 2 - Prob. 65CRCh. 2 - Prob. 66CRCh. 2 - Prob. 67CRCh. 2 - Prob. 68CRCh. 2 - Prob. 69CRCh. 2 - Prob. 70CRCh. 2 - Prob. 71CRCh. 2 - Prob. 72CRCh. 2 - Prob. 73CRCh. 2 - Prob. 74CRCh. 2 - Prob. 75CRCh. 2 - Prob. 76CRCh. 2 - Prob. 77CRCh. 2 - Prob. 78CRCh. 2 - Prob. 79CRCh. 2 - Prob. 80CRCh. 2 - Prob. 81CRCh. 2 - Prob. 82CRCh. 2 - Prob. 83CRCh. 2 - Prob. 84CRCh. 2 - Prob. 85CRCh. 2 - Prob. 86CRCh. 2 - Prob. 87CRCh. 2 - Prob. 88CRCh. 2 - Prob. 89CRCh. 2 - Prob. 90CRCh. 2 - Prob. 91CRCh. 2 - Prob. 92CRCh. 2 - Prob. 93CRCh. 2 - Prob. 94CRCh. 2 - Prob. 95CRCh. 2 - Prob. 96CRCh. 2 - Prob. 97CRCh. 2 - Prob. 98CRCh. 2 - Prob. 99CRCh. 2 - Prob. 100CRCh. 2 - Prob. 101CRCh. 2 - Prob. 102CRCh. 2 - Prob. 103CRCh. 2 - Prob. 104CRCh. 2 - Prob. 105CRCh. 2 - Prob. 106CRCh. 2 - Prob. 107CRCh. 2 - Prob. 108CRCh. 2 - Prob. 109CRCh. 2 - Prob. 110CRCh. 2 - Prob. 111CRCh. 2 - Prob. 112CRCh. 2 - Prob. 113CRCh. 2 - Prob. 114CRCh. 2 - Prob. 115CRCh. 2 - Prob. 116CRCh. 2 - Prob. 117CRCh. 2 - Prob. 118CRCh. 2 - Prob. 119CRCh. 2 - Prob. 120CRCh. 2 - Prob. 121CRCh. 2 - Prob. 122CRCh. 2 - Prob. 123CRCh. 2 - Prob. 124CRCh. 2 - Prob. 125CRCh. 2 - Prob. 126CRCh. 2 - Prob. 127CRCh. 2 - Prob. 128CRCh. 2 - Prob. 129CRCh. 2 - Prob. 130CRCh. 2 - Prob. 131CRCh. 2 - Prob. 132CRCh. 2 - Prob. 133CRCh. 2 - Prob. 134CRCh. 2 - Prob. 135CRCh. 2 - Prob. 136CRCh. 2 - Prob. 137CRCh. 2 - Prob. 138CRCh. 2 - Prob. 139CRCh. 2 - Prob. 140CRCh. 2 - Prob. 141CRCh. 2 - Prob. 142CRCh. 2 - Prob. 143CRCh. 2 - Prob. 144CRCh. 2 - Prob. 145CRCh. 2 - Prob. 146CRCh. 2 - Prob. 147CRCh. 2 - Prob. 148CRCh. 2 - Prob. 149CRCh. 2 - Prob. 150CRCh. 2 - Prob. 151CRCh. 2 - Prob. 152CRCh. 2 - Prob. 153CRCh. 2 - Prob. 154CRCh. 2 - Prob. 155CRCh. 2 - Prob. 156CRCh. 2 - Prob. 157CRCh. 2 - Prob. 158CRCh. 2 - Prob. 159CRCh. 2 - Prob. 160CRCh. 2 - Prob. 161CRCh. 2 - Prob. 162CRCh. 2 - Prob. 163CRCh. 2 - Prob. 164CRCh. 2 - Prob. 165CRCh. 2 - Prob. 166CRCh. 2 - Prob. 1CTCh. 2 - Prob. 2CTCh. 2 - Prob. 3CTCh. 2 - Prob. 4CTCh. 2 - Prob. 5CTCh. 2 - Prob. 6CTCh. 2 - Prob. 7CTCh. 2 - Prob. 8CTCh. 2 - Prob. 9CTCh. 2 - Prob. 10CTCh. 2 - Prob. 11CTCh. 2 - Prob. 12CTCh. 2 - Prob. 13CTCh. 2 - Prob. 14CTCh. 2 - Prob. 15CTCh. 2 - Prob. 16CTCh. 2 - Prob. 17CTCh. 2 - Prob. 18CTCh. 2 - Prob. 19CTCh. 2 - Prob. 20CTCh. 2 - Prob. 21CT
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