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

Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter6: Fractions
Section6.4: Least Common Denominators
Problem 13E
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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
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
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