Question

I need help in simplifying (3x^2 +3x -36) / (x^2-x-6) and how do you get the restrictions? I've gotten the wrong answer 3 times already :/

Answer 1

Restriction is when the denominator is 0.

Answer 2

So, \( \color{Black}{\Rightarrow x^2 - x - 6 \ne 0 }\)
Just solve it like a quadratic equation :)

Answer 3

wait huh?

Answer 4

Do you know how to solve quadratic equations?

Answer 5

i think so..

Answer 6

its with that formula right?

Answer 7

Yeah, many ways to solve.

Answer 8

If you know the formula, then you can use that :)

Answer 9

but how would that give you a restriction
like what is that

Answer 10

Well, the denominator must equal 0 right?

Answer 11

to get the restriction, the denominator must equal zero.

Answer 12

and after it equals zero...

Answer 13

could you show me an example or something?

Answer 14

Ok

Answer 15

\( \color{Black}{\Rightarrow \Large{4 \over x} }\)
x can't be 0 here.

Answer 16

so the restriction cant be 4?

Answer 17

Restriction on any value is when the denominator is 0.

Answer 18

To find the restriction, you must equate the denominator with 0.

Answer 19

maybe im just stupid but i dont get that at all, im sorry, thank you for trying though

Answer 20

lol no, it's my duty to explain

Answer 21

Anything over 0 is undefined.

Answer 22

\( \color{Black}{\Rightarrow \Large {y \over x - 4} }\)

1) equate the denominator with 0.

\( \color{Black}{\Rightarrow x - 4 = 0 }\)

2) solve for x.

\( \color{Black}{\Rightarrow x = 4 }\)

Answer 23

So, 4 is the restriction. This was an example tho.

Answer 24

could you work out the problem above to show me ? and like explain as you go? cause the one above seems alot more complicated thant he examples

Answer 25

Okay. As I told you, equate the denominator with 0.

\( \color{Black}{\Rightarrow x^2 - x - 6 = 0}\)

Answer 26

Use the formula now.

\( \color{Black}{\Rightarrow x = \Large {-b \pm \sqrt{b^2 - 4ac} \over 2a} }\)

Answer 27

a = 1
b = 1
c = -6

Answer 28

but dont you have to simplify (3x^2 +3x -36) / (x^2-x-6) before you can even get the restriction?

Answer 29

No, you don't have to :)

Answer 30

Well you can but it'd make it longer

Answer 31

but what if the problem tells you to :(

Answer 32

Then simplify it....

Answer 33

You can simplify by factoring.

Answer 34

thats part of what i dont get.;///;.......

Answer 35

Factor the numerator and denominator first.

Answer 36

-36 doesnt factor to equal +3 though for the numerator

Answer 37

3 is common in the numerator :D

Answer 38

so then how would you right that

Answer 39

write**

Answer 40

\( \color{Black}{\Rightarrow 3(x^2 + x - 12) }\)

Answer 41

^ you can factor this

Answer 42

ohhhh ya i forgot about that way

Answer 43

wait you can factor what

Answer 44

\( \color{Black}{\Rightarrow 3(x^2 -3x + 4x -12) }\)
:D

Answer 45

oh okok

Answer 46

so 3((x-3)(x+4)-12) or something i forget

Answer 47

Yeah... \(3(x + 3)(x - 4)\)

Answer 48

And now factor the denominator:

\( \color{Black}{\Rightarrow x^2 - 3x + 2x - 6}\)

Answer 49

(x-3)(x+2)

Answer 50

yep

Answer 51

so that's it... simplified.

Answer 52

Now as you know factoring, no need to use the formula

Answer 53

but you still hae to divide the two right

Answer 54

\( \color{Black}{\Rightarrow (x - 3)(x + 2) = 0 }\)
Can you solve now? this will get the restrictions

Answer 55

are the restrictions the roots of like a graph

Answer 56

yes.

Answer 57

wowwwwwww

Answer 58

hahahahah i am so sorry

Answer 59

but wait 3(x+3)(x+4) / (x-3)(x+2)

Answer 60

Yeah, but still this can't simplify

Answer 61

so then how would you even find the roots??

Answer 62

Use the zero product rule

Answer 63

the what

Answer 64

\( \color{Black}{\Rightarrow (x -3) = 0 }\)

\( \color{Black}{\Rightarrow (x + 4) = 0 }\)

This will get the roots that'd be the restrictions

Answer 65

does the problem have to simplify to look like that for you to be able to do that?

Answer 66

Hmm?

Answer 67

like in order to get the roots the problem usually has to simplify to look like ⇒(x−3)=0

⇒(x+4)=0 that to be able to equal it to zero

Answer 68

You have to solve these equations to get the solutions.

Answer 69

well ya which would be +3 and -4 but

Answer 70

Yeah, those ARE the restrictions to x :)

Answer 71

if you cant simplify 3(x+3)(x-4) / (x-3)(x+2) anymore than what makes the (x-3)and (x+4) part so special, wouldnt the other parts like (x+2) =0 to get -2 as a restriction too?

Answer 72

No! No!

Answer 73

The fraction is undefined when the denominator is 0, not the numerator.

Answer 74

but the denominator isnt zero.......

Answer 75

The denominator cannot equal 0.

Answer 76

ohh ever?

Answer 77

It can never equal zero lol zero is a bad number

Answer 78

the numerator can be zero on the other hand

Answer 79

oh ya cause then the top would equal zero too

Answer 80

wait....

Answer 81

the numerator is still 3(x+3) (x-4) though.... what happened to thee three

Answer 82

if you divide zero by something, it's zero.
if you divide something by zero, YOU JUST CANT

Answer 83

hello?

Answer 84

hi?

Answer 85

the numerator is still 3(x+3) (x-4) though.... what happened to the three

Answer 86

i get that the restrictions came from (x+3)(x-4) but in the numerator there is a three in from of the two

Answer 87

does that just dissapear?

Answer 88

Hmm?

Answer 89

i still dont get it

Answer 90

Lol jo wil explain

Answer 91

If you don't mind, can I start from the beginning?

Answer 92

so the answer is 3(x+3)(x-4) / (x-3)(x+2) with restrictions of -3 and +4
as long as its right you guys dont have to worry about explaning it haha

Answer 93

actually thatd be awesome as long as you dont mind.....

Answer 94

GIVE JO THE MEDALS

Answer 95

Nope :|
Do you mind re-post the question? Lag here :|

Answer 96

Then I deserve the medals :P

Answer 97

Take back the medals given to me!!!

Answer 98

Done