Multiply rational expressions

G

Guest

Guest
I got some of these ok, but this one I didn't. Could I get some help?

w^2 - 4w w-4
___________ . ___________
w^2-8w + 16 w^2 + w


w(w-4) . w-4 1
______ ________ = _________
? w(w+1)


This is all I have broke down so far :?
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,442
afreemanny said:
I got some of these ok, but this one I didn't. Could I get some help?
w^2 - 4w w-4
___________ . ___________
w^2-8w + 16 w^2 + w

w(w-4) . w-4 1
______ ________ = _________
? w(w+1)

This is all I have broke down so far :?
You was agettin' there, mom!

w^2 - 8w + 16 = (w - 4)(w-4)
SO:
w(w - 4)(w - 4)
==============
(w - 4)(w - 4)w(w + 1)

look at all them (w-4)'s; cancel 'em out:
w
======
w(w + 1)

now shoot the 2 w's:
1
====
w + 1
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, afreemanny!

Denis is absolutely correct . . . here's my version.

You factored everything nicely . . . except the trinomial.
. . Are you fuzzy on that procedure?

. . . w<sup>2</sup> - 4w . . . . w - 4
. --------------- . -----------
. w<sup>2</sup>- 8w + 16 . . w<sup>2</sup> + w
.
. . . . . . . . . . . . . . . . . .w(w - 4) . . . .w - 4
Factor everything: . --------------- . ------------
. . . . . . . . . . . . . . . (w - 4)(w - 4) . w(w + 1)

Then cancel factors common to the numerator and the denominator.
[I'm sure you're familiar with this part, though]

. . There is a "single w" on top and another on the bottom.
. . There is a (w - 4) on top and another on the bottom.
. . And there is <u>another</u> pair of (w - 4)'s to cancel.

. . . . . . . . . . 1
Answer: . --------
. . . . . . . . w + 1
 
G

Guest

Guest
Thanks

Thanks so much ! I get stuck on the dardest things!
 
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