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NP Chapter 5 Exponent Strategy Page 104 Factoring

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ray_serrano
 Post subject: NP Chapter 5 Exponent Strategy Page 104 Factoring
 Post Posted: Thu Jun 04, 2009 5:55 am 
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Course Students


Posts: 10
I don’t understand how does a^b(1 – a^-1) becomes a^b-1(a – 1), and how does p(q + r) + s(q + r) becomes (p + s)(q + r)? Please also provide where is this explained in the guides.
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farooq.mazhar
 Post subject: Re: NP Chapter 5 Exponent Strategy Page 104 Factoring
 Post Posted: Sat Jun 06, 2009 9:56 am 
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Students


Posts: 6
ray_serrano wrote:
I don’t understand how does a^b(1 – a^-1) becomes a^b-1(a – 1), and how does p(q + r) + s(q + r) becomes (p + s)(q + r)? Please also provide where is this explained in the guides.




p(q + r) + s(q + r).
suppose (q+r) = X.
pX + sX.

X(p + s). Now put the value of X.
(q+r)(p+s).

a^b(1 – a^-1).

We can write 1/a= a^-1.

So a^b(1-1/a)
= a^b[(a-1)/a]
= a^b(a-1) . (1/a)
= a^b(1/a). (a-1)
= a^b . a^-1 (a-1)
When base is common, add the powers.
= a^(b+(-1)) . (a-1)
= a^(b-1) (a-1)
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esledge
 Post subject: Re: NP Chapter 5 Exponent Strategy Page 104 Factoring
 Post Posted: Tue Jul 21, 2009 11:54 am 
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ManhattanGMAT Staff


Posts: 901
Location: St. Louis, MO
Good job Farooq!

ray_serrano wrote:
I don’t understand how does a^b(1 – a^-1) becomes a^b-1(a – 1)...

Just for clarity, I'll add some parentheses to the question and to Farooq's explanation. It's tough to show complex exponents and fractions accurately online without superscripts and fraction formatting:

Why does [a^b]*[1 - a^(-1)] = [a^(b - 1)]*[a - 1]?

We can write a^(-1) as 1/a.

So [a^b]*[1 - a^(-1)] = [a^b]*[1 - (1/a)]
= [a^b]*[(a/a) - (1/a)]
= [a^b]*[(a-1)/a]
= [a^b]*[(1/a)*(a-1)]
= [a^b]*[(a^(-1)]*[a - 1]
When base is common (moderator note: as it is in the first two terms of this product), add the powers.
= [a^(b + (-1))]*[a - 1]
= [a^(b - 1)]*[a - 1]

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Emily Sledge
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ManhattanGMAT
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