Question

@UsukiDoll

Answer 1

hiiiii

Answer 2

hey :D

Answer 3

okay first question

Answer 4

heres the first 3

Answer 5

ok can you hold a minute I'm answering another question

Answer 6

yes I can

Answer 7

k back

Answer 8

so GCF is the greatest common factor.. what doe the equation have in common

Answer 9

so what does the equation \[6y^2+12xy^2 \] have in common? hint: there's a variable and a number in common

Answer 10

umm this is for 10?

Answer 11

yes

Answer 12

they both have 6 and Y^2

Answer 13

yes they do

Answer 14

so when there is something in common we can place that common part to the side.. so 6y^2 ( whatever there is left)

Answer 15

so since a 1 and 2x is left behind we have
6y^2(1+2x)
we can check easily by distributing them back
6y^2+12xy^2

Answer 16

so it A?

Answer 17

yes

Answer 18

for #11.. there is only a number in common because there are 2 terms with a variable but the last one is just a number

Answer 19

3 is a common factor

Answer 20

yes :D

Answer 21

so 3 is on the side and what do we have left ( ) ?

Answer 22

its B?

Answer 23

no because originally we have 3 terms.. B only has two terms.

Answer 24

there's a way to make this a bit faster... since 3 is the common factor we just divide 12/3 18/3 and 3/3

Answer 25

Got you.. so its D?

Answer 26

yes

Answer 27

for twelve, i have to right the answer out.

Answer 28

hmm there is just a number is common.. it's similar to the previous problem

Answer 29

okay.

Answer 30

so what number can be used to divide 3, 9, and 27 ?

Answer 31

3

Answer 32

is my answer ...

Answer 33

3(x2-3x+9) ?

Answer 34

wow you're picking this up fast that's right \[3(x^2-3x+9) \rightarrow 3x^2-9x+27\]

Answer 35

okay next 3

Answer 36

is 13 d?

Answer 37

just a quick guess :/

Answer 38

wait no I mean b

Answer 39

yes it is B for 13

Answer 40

14 b also?

Answer 41

yes

Answer 42

okay 15 now

Answer 43

2 is common

Answer 44

yes 2 is common

Answer 45

2(x2+9x+8)

Answer 46

correct

Answer 47

but I have to do the +C -C

Answer 48

hmmm that would mean that x^2+9x+8 is factorable

Answer 49

2(x+8)(x+1)

Answer 50

so there's more than just factoring out the 2 .... SOOO FAST! :D

Answer 51

did I get it right?

Answer 52

yeah

Answer 53

yay! next 3

Answer 54

well 2 lol

Answer 55

we need long division for this xD

Answer 56

what does that mean?

Answer 57

I don't understand :/

Answer 58

I'm getting burned out.. it's like .. oh mai brain T_T

Answer 59

lol geeeeez louiseeee

Answer 60

wait. see now I'm not thinking right... we just use factoring and take out the like terms

Answer 61

the choices gave me a hint... if it was long division the answer would look different.. ok I can do this now xD!

Answer 62

for the first problem, there is a perfect square binomial

Answer 63

Yay we are back on the road lol and okay.

Answer 64

so x^2-16... that's a perfect square in the form of (x^2-y^2) = (x+y)(x-y)... so what is the square root of 16?

Answer 65

4?

Answer 66

yes so we have (x+4)(x-4) in the denominator.

Answer 67

that makes it A?

Answer 68

I think we got lucky on the numerator part of this since we need to find a combination of numbers that can produce an 8 and a 16

4 x 4 = 16
4+4 is also 8 and since the pattern is all plus signs, our end result should be something like (x+4)(x+4) and wow it is A

Answer 69

because there is at least one pair of x+4 (both in numerator and denominator) in common so that gets cancelled.

Answer 70

okay 17 now

Answer 71

numerator is that there is one number in common
denominator is that it can be split up since 12 is 6 x 2 = 12 and 6-2 = 4 the problem is should the 2 be negative or the 6 be negative in order to achieve the result of x^2+4x-12

Answer 72

(x+6)(x-2)

Answer 73

yes an for the numerator?

Answer 74

5

Answer 75

yes 5 is common

5(x-2)/ (x+6)(x-2) who gets canceled out?

Answer 76

(x-2)

Answer 77

yes so the answer is 5/(x+6)

Answer 78

d?

Answer 79

I have to go for a bit... xD need a break xD!

Answer 80

okay

Answer 81

You there?

Answer 82

hello?

Answer 83

I have to go eat dinner. I'll be back

Answer 84

okay :)