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(1-p)(1-p¹+p²+p³+p⁴+p⁵+p⁶)

Guest Oct 11, 2017

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Guest · Answer 1 · 2017-10-11T10:09:17

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

AdamTaurus · Answer 2 · 2017-10-11T11:25:04

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

Guest · Answer 3 · 2017-10-11T11:43:42

#3

0

Did you get it right?

Guest Oct 11, 2017

#4

+895

+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

CPhill · Answer 4 · 2017-10-12T04:47:03

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

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