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(1-p)(1-p¹+p²+p³+p⁴+p⁵+p⁶)

Guest Oct 11, 2017
 #1
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0

Expand the following:
(1 - p) (p^6 + p^5 + p^4 + p^3 + p^2 - p^1 + 1)

 | | | | | | | | | | | | -p | + | 1
 | | p^6 | + | p^5 | + | p^4 | + | p^3 | + | p^2 | - | p | + | 1
 | | | | | | | | | | | | -p | + | 1
 | | | | | | | | | | p^2 | - | p | + | 0
 | | | | | | | | -p^3 | + | p^2 | + | 0 | + | 0
 | | | | | | -p^4 | + | p^3 | + | 0 | + | 0 | + | 0
 | | | | -p^5 | + | p^4 | + | 0 | + | 0 | + | 0 | + | 0
 | | -p^6 | + | p^5 | + | 0 | + | 0 | + | 0 | + | 0 | + | 0
-p^7 | + | p^6 | + | 0 | + | 0 | + | 0 | + | 0 | + | 0 | + | 0
-p^7 | + | 0 | + | 0 | + | 0 | + | 0 | + | 2 p^2 | - | 2 p | + | 1:
-p^7 + 2 p^2 - 2 p + 1

Guest Oct 11, 2017
 #2
avatar+771 
+1

Hey, this was on my PSAT test that I took today.....

AdamTaurus  Oct 11, 2017
 #3
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0

Did you get it right?

Guest Oct 11, 2017
 #4
avatar+771 
+1

I'm not sure. I did it differently. I took (1-p) and distributed it into (1-p1+p2+p3+p4+p5+p6).

I got: \((1-p^1+p^2+p^3+p^4+p^5+p^6)+(-p^1+p^2-p^3-p^4-p^5-p^6-p^7)\)

This is the same as: \(1-p^1+p^2+p^3+p^4+p^5+p^6+-p^1+p^2-p^3-p^4-p^5-p^6-p^7\)

I just used parenthesis for clarity.

I then combined like terms.

\(1-2p^1+2p^2-p^7\)

2p1 is the same as 2p.

1-2p+2p2-p7

AdamTaurus  Oct 12, 2017
 #5
avatar+93038 
+1

Note

 

(1 - p)  ( 1 - p + p^2 + p^3 + p^4 + p^5 + p^6)  =

 

( 1 - p + p^2 + p^3 + p^4 + p^5 + p^6)

(     -p + p^2 -  p^3 - p^4  - p^5  - P^6  - p^7  )  =

 

1 - 2p  + 2p^2  - p^7   

 

 

cool cool cool   

CPhill  Oct 12, 2017

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