Explanation Multiplication of two binomial expression, expressed the terms as sum or difference is carried by the multiplication of each term of the first binomial expression with second binomial expression. Calculation: Given, 3 x 3 ( a + 3 ) 3 − 28 x ( a + 3 ) 3 + 25 ( a + 3 ) 3 Factor out the common factor ( a + 3 ) 3 ( a + 3 ) 3 ( 3 x 2 − 28 x + 25 ) Complete factor Taking, 3 x 2 − 28 x + 25 Factor of 3 x 2 3 x 2 = 3 x ⋅ x (i.e.) 3 x 2 − 28 x + 25 = ( 3 x + □ ) ( x + □ ) Since, the middle term − 28 x has a positive numerical coefficient, then we factor 25 into positive factor. Factor of 25 25 = 25 ⋅ 1 , 25 = − 25 ⋅ − 1 Let us take factor 25 and 1 = ( 3 x + 25 ) ( x + 1 ) By using the FOIL method, = ( 3 x ⋅ x ) + ( 3 x ⋅ 1 ) + ( 25 ⋅ x ) + ( 25 ⋅ 1 ) Since multiplication is associative, = 3 ( x ⋅ x ) + ( 3 ⋅ 1 ) x + 25 ( x ) + ( 1 ⋅ 25 ) = 3 x 2 + 3 x + 25 x + 25 Combine the like term, = 3 x 2 + 28 x + 25 Here, 28 x is the incorrect middle term, Also, let us take factor − 25 and <

Prealgebra & Introductory Algebra (5th Edition) (What's New in Developmental Math)

5th Edition

ISBN: 9780134707631

Author: Elayn Martin-Gay

Publisher: PEARSON

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Chapter 11.3, Problem 98E

To determine

To factor:

3x3(a+3)3−28x(a+3)3+25(a+3)3

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Chapter 11.3, Problem 97E

Chapter 11.3, Problem 99E

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Chapter 11 Solutions

Prealgebra & Introductory Algebra (5th Edition) (What's New in Developmental Math)

Ch. 11.1 - Prob. 1CC Ch. 11.1 - Prob. 2CC Ch. 11.1 - Prob. 1P Ch. 11.1 - Prob. 2P Ch. 11.1 - Prob. 3P Ch. 11.1 - Prob. 4P Ch. 11.1 - Prob. 5P Ch. 11.1 - Prob. 6P Ch. 11.1 - Prob. 7P Ch. 11.1 - Prob. 8P

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To factor: 3 x 3 ( a + 3 ) 3 − 28 x ( a + 3 ) 3 + 25 ( a + 3 ) 3

To factor: 3x3(a+3)3−28x(a+3)3+25(a+3)3

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Chapter 11 Solutions

To factor:

3x3(a+3)3−28x(a+3)3+25(a+3)3