Expand using the binomial theorem: (2x2-y)4



To expand (A + B)n by the binomial theorem



Start out with the 1st term, which is 1AnB0



then:



1. Multiply the numerical coefficient by the exponent of A,

2. Divide that by the number of term

3. Write that number down to start the next term

4. Beside that write A with an exponent that is 1 less

5. Beside that write B with an exponent that is 1 more.



This will be the next term.  Then repeat the process with

the new term. You are finished when the

exponent of A becomes 0 and the exponent of B becomes n



To expand (2x2 - y)4



Start out with the 1st term followed by a + sign.



So we begin with 

 

1(2x2)4(-y)0 +



1. Multiply the numerical coefficient 1 by the exponent of (2x2),

   which is 4.  That gives 1×4 or 4

2. Divide that by the number of term

   The number of term is 1, since it's the 1st term, so we divide

   4 by 1 and get 4



3. Write that number down to start the next term

   So we have



1(2x2)4(-y)0 + 4



4. Beside that write (2x2) with an exponent that is 1 less than 4

   So we now have 

1(2x2)4(-y)0 + 4(2x2)3



5. Beside that write (-y) with an exponent that is 1 more than 0

   So we now have 



1(2x2)4(-y)0 + 4(2x2)3(-y)1



The exponent of (2x2) has not reached 0 so we put a + sign

repeat the process



1(2x2)4(-y)0 + 4(2x2)3(-y)1 +



No we repeat the process:



1. Multiply the numerical coefficient 4 by the exponent of (2x2),

   which is 3.  That gives 4×3 or 12

2. Divide that by the number of term

   The number of term is 2, since it's the 2nd term, so we divide

   12 by 2 and get 6



3. Write that number down to start the next term

   So we have



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6



4. Beside that write (2x2) with an exponent that is 1 less than 3

   So we now have 



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2



5. Beside that write (-y) with an exponent that is 1 more than 1

   So we now have 



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2(-y)2 



The exponent of (2x2) has not reached 0 so we put a + sign

repeat the process



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2(-y)2 +



Now we repeat the process:



1. Multiply the numerical coefficient 6 by the exponent of (2x2),

   which is 2.  That gives 6×2 or 12

2. Divide that by the number of term

   The number of term is 3, since it's the 3rd term, so we divide

   12 by 3 and get 4



3. Write that number down to start the next term

   So we have



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2(-y)2 + 4



4. Beside that write (2x2) with an exponent that is 1 less than 2

   So we now have 



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2(-y)2 + 4(2x2)1



5. Beside that write (-y) with an exponent that is 1 more than 2

   So we now have 



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2(-y)2 + 4(2x2)1(-y)3



The exponent of (2x2) has not reached 0 so we put a + sign

repeat the process



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2(-y)2 + 4(2x2)1(-y)3 +



Now we repeat the process:



1. Multiply the numerical coefficient 4 by the exponent of (2x2),

   which is 1.  That gives 4×1 or 4

2. Divide that by the number of term

   The number of term is 4, since it's the 4th term, so we divide

   4 by 4 and get 1



3. Write that number down to start the next term

   So we have



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2(-y)2 + 4(2x2)1(-y)3 + 1



4. Beside that write (2x2) with an exponent that is 1 less than 1

   So we now have 



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2(-y)2 + 4(2x2)1(-y)3 + 1(2x2)0



5. Beside that write (-y) with an exponent that is 1 more than 3

   So we now have 



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2(-y)2 + 4(2x2)1(-y)3 + 1(2x2)0(-y)6



The exponent of (2x2) has now reached 0 so we are done except for

simplifying.  Remember that a number raised to the 0 power is 1.



1(2x2)4(-y)0 + 4(2x2)3(-y)1 + 6(2x2)2(-y)2 + 4(2x2)1(-y)3 + 1(2x2)0(-y)4



1(24x8)(1) + 4(23x6)(-y) + 6(22x4)y2 + 4(2x2)(-y3) + 1(1)y4



16x8 - 32x6y + 24x4y2 - 8x2y3 + y4 



Edwin 

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