Question

help with factoring please :)

Answer 1

\[6x^4y^2 + 12x^2y^3 -9x^3y^4\]

Answer 2

i can try but im not saying i will know the anwser lol

Answer 3

ok, so are there any vaiables common to every term?

Answer 4

lol its okay @samanthagreer

Answer 5

no @FibonacciChick666

Answer 6

ok by the way love the username and pic

Answer 7

thanks ^.^ @samanthagreer

Answer 8

by that I mean, can you take the same variable out of each term evenly, like if I had 5x^2y+3x, I could take an x out of each term and get x(5xy+3)

Answer 9

ohh lol yes we got x and y

Answer 10

sorry im not sure

Answer 11

its fine u r good :D @samanthagreer

Answer 12

ok, so if we took out one x and one y, what would we have?

Answer 13

\[6x^4y^2+12x^2y^3−9x^3y^4\]
\[=xy(?)\]

Answer 14

\[xy(6x^3y + 12xy^2 - 9x^2y^3)\]

Answer 15

? @FibonacciChick666

Answer 16

correct, so now, we repeat the process, can we take anything else out?

Answer 17

can we remove any other x or y, how about a common divisor among the coefficients?

Answer 18

\[3xy(2x^3y + 4xy^2 - 3x^2y^3)\]

Answer 19

@FibonacciChick666 did i do it right?

Answer 20

that is correct, but you can pull out more

Answer 21

no

Answer 22

you can

Answer 23

oh :/

Answer 24

it's ok, so inside the parentheses only what does every term have in common?

Answer 25

\[3xy(2x^3y+4xy^2−3x^2y^3)\]

Answer 26

(hint, look at the variables)

Answer 27

idk D:

Answer 28

ok, so just look at the x and y, is there one still in every term?

Answer 29

at least one**

Answer 30

yes 3xy(2x3y+4xy2−3x2y3)
2x^3y

Answer 31

Why did you restate 2x^3y?

Answer 32

but yes, there is

Answer 33

cuz y is to the 1st power

Answer 34

but why did you restate it when you have 3 terms-not just one?

Answer 35

but yes, y is to the first power

Answer 36

cuz u said "at least one"

Answer 37

ah, ok, so by at least one, I mean when you look at all three terms, is there at least 1 x and at least one y in each remaining?

Answer 38

ohh lol im sooo slow and 3xy(2x3y+4xy2−3x2y3) is that our final answer?

Answer 39

no, it is not, like I said before we can still factor further

Answer 40

that is why we need to look inside the parentheses and see what is common to all three terms

Answer 41

i would say x^2 but we also have x alone same with y

Answer 42

ok that is a good observation

Answer 43

you are correct it is not x^2, it is however the "x alone same with y"

Answer 44

so we are done right

Answer 45

not yet

Answer 46

you have an x and a y in each term, you gotta pull it out

Answer 47

i did 3xy(...)

Answer 48

you can still pull out more

Answer 49

if you pull out another xy, you just multiply that by what you have already pulled out

Answer 50

u mean like increase the power?

Answer 51

you would when you multiplied

Answer 52

so immagine you did not have the 3xy, what would you do with just:\[(2x^3y+4xy^2−3x^2y^3)\]

Answer 53

xy(..) ??

Answer 54

@FibonacciChick666

Answer 55

fill in the inside of the parentheses plaese

Answer 56

sorry \[xy(2x^2 +4y -3xy^2)\]

Answer 57

good, ok so now you have told me that:
\[xy(2x2+4y−3xy2)=(2x3y+4xy2−3x2y3)\] essentially

Answer 58

so why can't we multiply both sides by 3xy, won't they still be the same?

Answer 59

3x^2y^2 (2x^2 +4y -3xy^2)

Answer 60

\[3x^2y^2 (2x^2 +4y -3xy^2)\]

Answer 61

there we go

Answer 62

thanks very much u r awesome :D sorry for being such a mess ;/

Answer 63

lol it's cool, everyone is when they first learn this. Don't worry

Answer 64

lol okay :)