Holt Mcdougal Larson Algebra 2: Student Edition 2012
Holt Mcdougal Larson Algebra 2: Student Edition 2012
1st Edition
ISBN: 9780547647159
Author: HOLT MCDOUGAL
Publisher: HOLT MCDOUGAL
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Chapter 2.8, Problem 1MRPS
Solution

a.

To write: A polynomial equation that you can use to find the value of x .

The polynomial equation would be x3+6x2+5x180=0 .

Given information:

The volume of the rectangular prism shown is 180 cubic inches.

  Holt Mcdougal Larson Algebra 2: Student Edition 2012, Chapter 2.8, Problem 1MRPS , additional homework tip  1

Formula used:

Volume of rectangular box is length times width times the height of the box.

Calculation:

The volume of the box would be:

  V=xx+1x+5=xxx+5+1x+5=xx2+5x+x+5=xx2+6x+5=x3+6x2+5x

Since volume of the box is 180 cubic inches, so replace V with 180 as:

  180=x3+6x2+5x0=x3+6x2+5x180

Therefore, the required polynomial equation would be x3+6x2+5x180=0 .

b.

To identify: All the possible rational solutions of the equation in part (a).

The possible rational solutions of the equation would be ±1,±2,±3,±4,±5,±6,±9,±10,±12,±15,±20,±30,±36,±45,±60,±90,±180 .

Given information:

The volume of the rectangular prism shown is 180 cubic inches.

  Holt Mcdougal Larson Algebra 2: Student Edition 2012, Chapter 2.8, Problem 1MRPS , additional homework tip  2

Formula used:

As per Rational Zero Theorem, the possible rational zeros for any function with integer coefficients can be obtained by dividing factors of leading coefficient of highest degree term with factors of the constant term as:

  pq=Factor of constant term a0Factor of leading coefficient term an

Calculation:

In the equation x3+6x2+5x180=0 , the leading coefficient is 1 and the constant term is 180 .

Factors of 1 are ±1 and factors of 180 are ±1,±2,±3,±4,±5,±6,±9,±10,±12,±15,±20,±30,±36,±45,±60,±90,±180 .

  pq=±11,±21,±31,±41,±51,±61,±91,±101,±121,±151,±201,±301,±361,±451,±601,±901,±1801=±1,±2,±3,±4,±5,±6,±9,±10,±12,±15,±20,±30,±36,±45,±60,±90,±180

Therefore, the possible rational solutions of the equation would be ±1,±2,±3,±4,±5,±6,±9,±10,±12,±15,±20,±30,±36,±45,±60,±90,±180 .

c.

To use: The synthetic division to find a rational solution of the equation. Show that no other real solutions exist.

The only real solution of the equation would be x=4 .

Given information:

The volume of the rectangular prism shown is 180 cubic inches.

  Holt Mcdougal Larson Algebra 2: Student Edition 2012, Chapter 2.8, Problem 1MRPS , additional homework tip  3

Calculation:

By checking possible zeros in the given function, it can be seen that x=4 is a zero of the function. Now, use synthetic division as shown below:

  41 6 5 180 4 40 1801 10 450

Because x=4 is a zero of the function, so the function can be written fx=x4x2+10x+45 .

Check for the solution of the quadratic equation using the determinant formula b24ac :

  D=b24ac=1024145=100180=80

Since the determinant is a negative number, therefore, there will be no real solutions for the quadratic equation and the only solution would be x=4 .

d.

To determine: What are the dimensions of the prism?

The dimensions of the prism are 4 inches by 5 inches by 9 inches.

Given information:

The volume of the rectangular prism shown is 180 cubic inches.

  Holt Mcdougal Larson Algebra 2: Student Edition 2012, Chapter 2.8, Problem 1MRPS , additional homework tip  4

Calculation:

Using x=4 , the other sides would be:

  x+14+15x+54+59

Therefore, the dimensions of the prism are 4 inches by 5 inches by 9 inches.

Chapter 2 Solutions

Holt Mcdougal Larson Algebra 2: Student Edition 2012

Ch. 2.1 - Prob. 3ECh. 2.1 - Prob. 4ECh. 2.1 - Prob. 5ECh. 2.1 - Prob. 6ECh. 2.1 - Prob. 7ECh. 2.1 - Prob. 8ECh. 2.1 - Prob. 9ECh. 2.1 - Prob. 10ECh. 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. 49PSCh. 2.1 - Prob. 50PSCh. 2.1 - Prob. 51PSCh. 2.1 - Prob. 52PSCh. 2.1 - Prob. 53PSCh. 2.1 - Prob. 54PSCh. 2.1 - Prob. 1DCCh. 2.1 - 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Algebra
ISBN:9780135163078
Author:Michael Sullivan
Publisher:PEARSON
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Algebra
ISBN:9780980232776
Author:Gilbert Strang
Publisher:Wellesley-Cambridge Press
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ISBN:9780077836344
Author:Julie Miller, Donna Gerken
Publisher:McGraw-Hill Education
Interpreting Graphs of Quadratic Equations (GMAT/GRE/CAT/Bank PO/SSC CGL) | Don't Memorise; Author: Don't Memorise;https://www.youtube.com/watch?v=BHgewRcuoRM;License: Standard YouTube License, CC-BY
Solve a Trig Equation in Quadratic Form Using the Quadratic Formula (Cosine, 4 Solutions); Author: Mathispower4u;https://www.youtube.com/watch?v=N6jw_i74AVQ;License: Standard YouTube License, CC-BY