+0  
 
0
104
5
avatar

(1-p)(1-p¹+p²+p³+p⁴+p⁵+p⁶)

Guest Oct 11, 2017
Sort: 

5+0 Answers

 #1
avatar
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+648 
+1

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

AdamTaurus  Oct 11, 2017
 #3
avatar
0

Did you get it right?

Guest Oct 11, 2017
 #4
avatar+648 
+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+79827 
+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

23 Online Users

avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details