In the expression (x+y)', the exponent represents repeated multiplication, so the expression may be written as (x+ y)(x+ y)(x+ y). One way to perform the multiplication is to multiply (x+ y)(x+ y), simplify the result, and then multiply by (x+ y) again. 1. Use the steps described above to perform the multiplication: (x+ y)(x+ y)(x+ y). While the steps above worked well for (x+y)' = (x+y)(x+y)(x+y) and can be extended to (x+y)*, (x+ y)', (x+y)°, and so on, performing the multiplication in this manner is inefficient for these bigger exponents. We can, however, look for patterns that make the multiplications easier. Another way to think about the multiplication is to consider the sum of all possible products where each product is formed by taking one term from each factor in parentheses. The expansion below should help to 1 2 3 clarify what we mean by this. The numbers above the variables in xxyxr mean that the first x comes from the first factor of (x+ y), the y comes from the second factor of (x+ y), and the second x comes from the third factor of (x+ y). 1st factor 2nd factor 3rd factor (x+ y) = (x+ y)(x + y) (x + y) 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 = xXxXx+ xXrxy+ xxyxr+ xxyxy+ yxrxr+ yxrxy+ yxyxr+ yxyxy
In the expression (x+y)', the exponent represents repeated multiplication, so the expression may be written as (x+ y)(x+ y)(x+ y). One way to perform the multiplication is to multiply (x+ y)(x+ y), simplify the result, and then multiply by (x+ y) again. 1. Use the steps described above to perform the multiplication: (x+ y)(x+ y)(x+ y). While the steps above worked well for (x+y)' = (x+y)(x+y)(x+y) and can be extended to (x+y)*, (x+ y)', (x+y)°, and so on, performing the multiplication in this manner is inefficient for these bigger exponents. We can, however, look for patterns that make the multiplications easier. Another way to think about the multiplication is to consider the sum of all possible products where each product is formed by taking one term from each factor in parentheses. The expansion below should help to 1 2 3 clarify what we mean by this. The numbers above the variables in xxyxr mean that the first x comes from the first factor of (x+ y), the y comes from the second factor of (x+ y), and the second x comes from the third factor of (x+ y). 1st factor 2nd factor 3rd factor (x+ y) = (x+ y)(x + y) (x + y) 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 = xXxXx+ xXrxy+ xxyxr+ xxyxy+ yxrxr+ yxrxy+ yxyxr+ yxyxy
Mathematics For Machine Technology
8th Edition
ISBN:9781337798310
Author:Peterson, John.
Publisher:Peterson, John.
Chapter43: Introduction To Equations
Section: Chapter Questions
Problem 1A
Related questions
Question
To the person answering this, be on the look out for questions that follow this first question. Would appreciate it. Thank you.
question has multiple parts!
Transcribed Image Text:In the expression (x+ y)', the exponent represents repeated multiplication, so the expression may be written as
(x+ y)(x+ y)(x + y).
One way to perform the multiplication is to multiply (x+ y)(x+ y), simplify the result, and then multiply by (x+ y)
again.
1.
Use the steps described above to perform the multiplication: (x+ y)(x+ y)(x+ y).
While the steps above worked well for (x+ y)³ = (x+ y)(x+ y)(x+ y) and can be extended to (x+ y)“,
(x + y)', (x+y)°, and so on, performing the multiplication in this manner is inefficient for these bigger
exponents. We can, however, look for patterns that make the multiplications easier.
Another way to think about the multiplication is to consider the sum of all possible products where each
product is formed by taking one term from each factor in parentheses. The expansion below should help to
1 2 3
clarify what we mean by this. The numbers above the variables in xxyx mean that the first x comes from the
first factor of (x+ y), the y comes from the second factor of (x+ y), and the second x comes from the third
factor of (x+ y).
1st factor 2nd factor 3rd factor
(x+ y)' = (x+ y)(x+ y)(x + y)
1 2 3 1 2 3
= xxxXx+ xXxXy+ xxyxx+ xxyxy+ yxxxx+ yXxxy+ yxyxx+ yxyxy
1 2 3
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Holt Mcdougal Larson Pre-algebra: Student Edition…
Algebra
ISBN:
9780547587776
Author:
HOLT MCDOUGAL
Publisher:
HOLT MCDOUGAL
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Holt Mcdougal Larson Pre-algebra: Student Edition…
Algebra
ISBN:
9780547587776
Author:
HOLT MCDOUGAL
Publisher:
HOLT MCDOUGAL
Glencoe Algebra 1, Student Edition, 9780079039897…
Algebra
ISBN:
9780079039897
Author:
Carter
Publisher:
McGraw Hill
