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im waiting :P

\[5yz ^{2}-35yz+10y ^{2}z\]

5 yz ( 5z - 7 + 2y)

WOoOOOooooOooooOooooo :P

5yz(z - 7 + 2y)

wait...-.- its z not 5z inside the bracket

5 yz ( z - 7 + 2y) soorry

x(y-3) -4(y-3)

you want us to factor that?

do yu simplify it first

and ya factor it

its already factored tho: i guess you can use distributive property and then factor it as a whole

(y-3) (x-4)?

is tht the ans

my bad :P
yes it is

Go Charmi :D

^__^

lol any other ones :P

2mn+3mp-4n-6p

2n(m-2) + 3p (m-2)

(m-2) (2n+3p)

wait

yeah use grouping :O

im losing myself

so is it right?

aha

\[x^{2}-12x+20\]

= 2mn+3mp-4n-6p
= (2mn - 4n) + ( 3mp - 6p)
= 2n( m - 2) + 3p (m - 2)
= (2n + 3p) (m-2)

\[x ^{2} - 12x + 20 \]

yup

i didnt lose myself :P

yu need 2 numbers tht multiply to 20 and add to 6 right

two numbers that add to 12 and multiply to 20 *

not 6

oh my bad :P

lol dw we all make mistakes P:

im lost :/

on what yu did

oh ok :P

thanks :D

and r yu sure its not (x-2) (x-10)

or it doesnt matter wht way yu put it as

\[y^{2}+3y-18\]

any ideas of multiples

i was stuck
here:
y(y+6) -3 (y+6)

wud it be
(y+6) (y-3)

someone cheer for her
that's right
you had the answer the whole time <3

WOOOOO:P

buhh cant yu still divide it
cause theres a 3 and a 6?

or jus leave it like tht :P

lol i believe you can leave it like that :D

aha ok :P

\[x^{2}-18\]

shud i leave the middle number as 0

what do you think this one is

how wud yu do this
i forgot :$

(a - b) (a+ b)
the middle number is eliminated :)

(x-6) (x+6)?

no wait (x-6) (x+3)

i believe the middle number is missing
cuz you cant factor that

yu cant?

so yu dun do nytiin to it?

well no matter how much we factor it
it wont match the question after factoring

good job dada