Multiply: (a) (m + 2)(m – 2), (b) (3y + 4)(3y – 4), (c) (b -6 +), and (d) (tª – 6u)(tª + 6u). To find the product of each pair of binomials, we will use the special-product rule for the sum and difference of the same two terms. This approach is faster than using the FOIL method. (a) (m + 2) and (m - 2) are the sum and difference of the same two terms, m and 2. (m + 2)(m – 2) = m2 22 The square of the first term, m. The square of the second term, 2. = m2 -O (b) (3y + 4) and (3y - 4) are the sum and difference of the same two terms, 3y and 4. (3y + 4)(3y – 4) = (3y)? 42 The square of the first term, 3y. The square of the second term, 4. |y² - O %3D (c) By the commutative property of multiplication, the special-product rule can be written with the factor containing the – symbol first. That is (A – B)(A + B) = A² - B?. Since (o-금) and (o + 금) are the difference and sum of the same two terms, b and 2/3, we have (금)" +o %3D The square of the first term, b. The square of the second term, 2/3. = b2 -O (d) (tª - 6u) and (t + 6u) are the difference and sum of the same two terms, tª and 6u. (e4 – 6u)(t + 6u) = (4)2 (6u)² The square of the first term, tª. The square of the second term, 6u. = t8 - Multiply: (a) (b + 7)(b-7), (b) (2m + 8)(2m- 8), (c) (s-흥)(s + 흥), and (d) (c³ + 3d)(c³ - 3d). (a) (b) (c) (d)
Multiply: (a) (m + 2)(m – 2), (b) (3y + 4)(3y – 4), (c) (b -6 +), and (d) (tª – 6u)(tª + 6u). To find the product of each pair of binomials, we will use the special-product rule for the sum and difference of the same two terms. This approach is faster than using the FOIL method. (a) (m + 2) and (m - 2) are the sum and difference of the same two terms, m and 2. (m + 2)(m – 2) = m2 22 The square of the first term, m. The square of the second term, 2. = m2 -O (b) (3y + 4) and (3y - 4) are the sum and difference of the same two terms, 3y and 4. (3y + 4)(3y – 4) = (3y)? 42 The square of the first term, 3y. The square of the second term, 4. |y² - O %3D (c) By the commutative property of multiplication, the special-product rule can be written with the factor containing the – symbol first. That is (A – B)(A + B) = A² - B?. Since (o-금) and (o + 금) are the difference and sum of the same two terms, b and 2/3, we have (금)" +o %3D The square of the first term, b. The square of the second term, 2/3. = b2 -O (d) (tª - 6u) and (t + 6u) are the difference and sum of the same two terms, tª and 6u. (e4 – 6u)(t + 6u) = (4)2 (6u)² The square of the first term, tª. The square of the second term, 6u. = t8 - Multiply: (a) (b + 7)(b-7), (b) (2m + 8)(2m- 8), (c) (s-흥)(s + 흥), and (d) (c³ + 3d)(c³ - 3d). (a) (b) (c) (d)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
Related questions
Question
100%
![**Multiplying Binomials Using Special-Product Rule**
To find the product of each pair of binomials, we will use the special-product rule for the sum and difference of the same two terms. This approach is faster than using the FOIL method.
### Examples:
1. **(a) (m + 2)(m - 2)**
- \( (m + 2) \) and \( (m - 2) \) are the sum and difference of the same two terms, \( m \) and \( 2 \).
- \[
(m + 2)(m - 2) = m^2 - 2^2
\]
- The square of the first term, \( m \).
- The square of the second term, \( 2 \).
- \[
= m^2 - 4
\]
2. **(b) (3y + 4)(3y - 4)**
- \( (3y + 4) \) and \( (3y - 4) \) are the sum and difference of the same two terms, \( 3y \) and \( 4 \).
- \[
(3y + 4)(3y - 4) = (3y)^2 - 4^2
\]
- The square of the first term, \( 3y \).
- The square of the second term, \( 4 \).
- \[
= 9y^2 - 16
\]
3. **(c) \( \left( b - \frac{2}{3} \right) \left( b + \frac{2}{3} \right) \)**
- By the commutative property of multiplication, the special-product rule can be written with the factor containing the \(-\) symbol first. That is, \((A - B)(A + B) = A^2 - B^2\).
- Since \( \left( b - \frac{2}{3} \right) \) and \( \left( b + \frac{2}{3} \right) \) are the difference and sum of the same two terms, \( b \) and \( \frac{2}{3} \), we have:
- \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4c02ecae-9f3c-4011-90ce-878cb1c02c8e%2F1b416897-2f74-4c49-a83c-2dccb2127e09%2Ftd5bi4_processed.png&w=3840&q=75)
Transcribed Image Text:**Multiplying Binomials Using Special-Product Rule**
To find the product of each pair of binomials, we will use the special-product rule for the sum and difference of the same two terms. This approach is faster than using the FOIL method.
### Examples:
1. **(a) (m + 2)(m - 2)**
- \( (m + 2) \) and \( (m - 2) \) are the sum and difference of the same two terms, \( m \) and \( 2 \).
- \[
(m + 2)(m - 2) = m^2 - 2^2
\]
- The square of the first term, \( m \).
- The square of the second term, \( 2 \).
- \[
= m^2 - 4
\]
2. **(b) (3y + 4)(3y - 4)**
- \( (3y + 4) \) and \( (3y - 4) \) are the sum and difference of the same two terms, \( 3y \) and \( 4 \).
- \[
(3y + 4)(3y - 4) = (3y)^2 - 4^2
\]
- The square of the first term, \( 3y \).
- The square of the second term, \( 4 \).
- \[
= 9y^2 - 16
\]
3. **(c) \( \left( b - \frac{2}{3} \right) \left( b + \frac{2}{3} \right) \)**
- By the commutative property of multiplication, the special-product rule can be written with the factor containing the \(-\) symbol first. That is, \((A - B)(A + B) = A^2 - B^2\).
- Since \( \left( b - \frac{2}{3} \right) \) and \( \left( b + \frac{2}{3} \right) \) are the difference and sum of the same two terms, \( b \) and \( \frac{2}{3} \), we have:
- \
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps with 21 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, algebra and related others by exploring similar questions and additional content below.Recommended textbooks for you
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:
9780134463216
Author:
Robert F. Blitzer
Publisher:
PEARSON
Contemporary Abstract Algebra
Algebra
ISBN:
9781305657960
Author:
Joseph Gallian
Publisher:
Cengage Learning
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra and Trigonometry (6th Edition)
Algebra
ISBN:
9780134463216
Author:
Robert F. Blitzer
Publisher:
PEARSON
Contemporary Abstract Algebra
Algebra
ISBN:
9781305657960
Author:
Joseph Gallian
Publisher:
Cengage Learning
Linear Algebra: A Modern Introduction
Algebra
ISBN:
9781285463247
Author:
David Poole
Publisher:
Cengage Learning
Algebra And Trigonometry (11th Edition)
Algebra
ISBN:
9780135163078
Author:
Michael Sullivan
Publisher:
PEARSON
Introduction to Linear Algebra, Fifth Edition
Algebra
ISBN:
9780980232776
Author:
Gilbert Strang
Publisher:
Wellesley-Cambridge Press
College Algebra (Collegiate Math)
Algebra
ISBN:
9780077836344
Author:
Julie Miller, Donna Gerken
Publisher:
McGraw-Hill Education