how to factor when theres 4 different things?soconfused

perform the indicated operation. answer in lowest terms......


5a-5b-a^2+b^2 4a^2-b^2
------------------ x ---------------
4a^2-4ab+b^2 2a^2-1ab-b^2






so i totally understand that like you get an answer that looks like this


( ) ( ) ( ) ( )
----------------------------------- ------------------------------
( ) ( ) ( ) ( )





what i dont know is how to factor the first one because there are 4 different things and i didnt learn that and then what goes in between the two fractions? is it a multiplication or a plus sign?

2 2
5a-5b-a +b

(5 ) (a ) thats all i know

really lost said:

5a-5b-a^2+b^2

Click to expand...

Sometimes you just have to play around and see what you find.

Factor pieces.

5(a-b) - a^2 + b^2

Hmmmm...add or subtract parentheses.

5(a-b) - (a^2-b^2)

Aha!!! Something jumped out at me.

5(a-b) - (a+b)(a-b) = [5 - (a+b)](a-b) = (5-a-b)(a-b)

If that hadn't worked, I may have started shuffling things around.

Disagree TK; I get (5 - a - b) / (2a - b)
(and tested it!)

I'm glad you disagree, since I did only the first numerator.

R dee R dee R R R

remember who laughed like that?

Hello, really lost!

TK explained how we factor a polynomial with 4 (or more) terms.
It's called "factoring by grouping".

We have: . 5a - 5b - a² + b²

There is no common factor running through all four terms,
. . so we try factoring them "in pairs".

Take "5" out of the first two,, take "-" out the last two.
. . . 5(a - b) - (a² - b²)

Nothing's happening . . . but the last group is a "difference of squares" . . . factor it.
. . . 5(a - b) - (a + b)(a - b)

Now we see a common factor in the two groups: 5(a - b) - (a + b)(a - b)

Factor it out: . (a - b) (5 - [a + b]) .= .(a - b)(5 - a - b)

I assume you can factor the other polynomials . . .


. . . . . . . . . . . . . . . . . . . .(a - b)(5 - a - b) . . .(2a - b)(2a + b)
The problem becomes: . -------------------- .x .--------------------
. . . . . . . . . . . . . . . . . . . .(2a - b)(2a - b) . . . .(2a + b)(a - b)

. . . and now you can reduce it . . .


[It was Ralph Kramden, Denis . . . "Really funny, Alice ... Har-dee-har-dee-har-har!"]