Question

how do i get the LCM for this??
\[\frac{2x^2 - x - 1}{x^2 - 9} \times \frac{x+3}{2x + 1}\]

Answer 1

just factorise

Answer 2

oh...any hints?

Answer 3

You don't have to! In multiplication, you just multiply, no LCM required, or you could factorize and also I see that they can be cancelled!

Answer 4

yeah x^-9= x^2-3^2

Answer 5

then when do you LCM??

Answer 6

@lgbasallote You take the LCM in addition or subtraction.

Answer 7

oh i see..sorry

Answer 8

@lgbasallote 2x^2-x-1 = ?

Answer 9

uhmmm wait...i know this...you break down the middle term right?

Answer 10

Yes

Answer 11

i guess -2x + x could work?

Answer 12

Yes lgba, quadratic factoring.

Answer 13

Yep.

Answer 14

so 2x^2 - 2x + x + 1?

Answer 15

Yes, now factorize by grouping.

Answer 16

right

Answer 17

uhmmm 2(x^2 - x) + x + 1? this doesnt make sense o.O

Answer 18

take 2x out not just 2

Answer 19

ohhh finklesauce

Answer 20

2x(x+1) + x+ 1...that looks better

Answer 21

\(\Large \color{purple}{\rightarrow 2x(x - 1) }\)

Answer 22

i factor out x+1 this time right?

Answer 23

Do you know how to factor by grouping?

Answer 24

uhmm...idk

Answer 25

its 2x^2 - 2x + x - 1

Answer 26

\(\Large \color{purple}{\rightarrow 2x(x - 1) - (-x - 1) }\)

Answer 27

that still doesnt make sense o.O

Answer 28

i cant factor out x - 1....

Answer 29

@ParthKohli

Answer 30

Not sure about that :/

Answer 31

2x^2 - 2x + x - 1
2x(x-1) + 1(x-1)

Answer 32

ohhhh now that makes sense...

Answer 33

Oh yeah I forgot that there's a positive x :P

Answer 34

so x - 1 gets factored out....

(x-1)(2x -1)?

Answer 35

\(\Large \color{purple}{\rightarrow (2x + 1)(x - 1) }\)

Answer 36

oh plus

Answer 37

Yeah!! plus :D

Answer 38

so what do i do with that now =_=

Answer 39

I used to know a genius lgba ://

Answer 40

Cancel out with the denominator of the other fraction.

Answer 41

Put the factored terms in the fractions and see what you can cancel!

Answer 42

so \[\frac{x-1}{x^2 - 9} \times x +3\]?? is that right o.O

Answer 43

Yes!

Answer 44

oh goodie

Answer 45

You can also factorize x^2 - 9

Answer 46

how??

Answer 47

\(\Large \color{purple}{\rightarrow x^2 - 9 = x^2 - 3^2 = (x + 3)(x - 3) }\)

Answer 48

oh

Answer 49

x+ 3 gets canceled?

Answer 50

Now, you can cancel x + 3 too

Answer 51

Yes

Answer 52

hmm..im getting \[\frac{x-1}[x+3}\]

Answer 53

\[\frac{x-1}{x+3}\]

Answer 54

is that right?

Answer 55

How did you get that?

Answer 56

i dont know :/

Answer 57

you said to factor and cancel =_=

Answer 58

\(\Large \color{purple}{\rightarrow {a \over b} * {c \over d} = {ac \over cd} }\)

Uh huh?

Answer 59

???

Answer 60

ohhh oh yeah *facepalm*

Answer 61

\[\frac{\cancel{(2x+1)}(x-1)}{\cancel{(x+3)}(x-3)} \times \frac{\cancel{(x+3)}}{\cancel{(2x+1)}}\]

Answer 62

So., you were right ^-^