NP Chapter 5 Exponent Strategy Page 104 Factoring

[ray_serrano]

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.

[farooq.mazhar]

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)

[esledge]

Good job Farooq!

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]