Factor:

- anonymous

Factor:

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- anonymous

im waiting :P

- anonymous

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

- anonymous

5 yz ( 5z - 7 + 2y)

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## More answers

- anonymous

WOoOOOooooOooooOooooo :P

- anonymous

5yz(z - 7 + 2y)

- anonymous

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

- anonymous

5 yz ( z - 7 + 2y) soorry

- anonymous

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

- anonymous

you want us to factor that?

- anonymous

do yu simplify it first

- anonymous

and ya factor it

- anonymous

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

- anonymous

(y-3) (x-4)?

- anonymous

is tht the ans

- anonymous

my bad :P
yes it is

- anonymous

Go Charmi :D

- anonymous

^__^

- anonymous

lol any other ones :P

- anonymous

2mn+3mp-4n-6p

- anonymous

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

- anonymous

well for this one you have to find a variable thats common to all the coefficients
but i dont think its possible to factor this

- anonymous

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

- anonymous

wait

- anonymous

yeah use grouping :O

- anonymous

im losing myself

- anonymous

so is it right?

- anonymous

aha

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yup

- anonymous

\[x ^{2} - 10x - 2x + 20 \]\[(x ^{2} - 2x) + (-10x +20)\] \[x(x-2) -10 (x - 2)\] \[(x - 10) (x - 2)\]

- anonymous

i didnt lose myself :P

- anonymous

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

- anonymous

two numbers that add to 12 and multiply to 20 *

- anonymous

not 6

- anonymous

oh my bad :P

- anonymous

lol dw we all make mistakes P:

- anonymous

im lost :/

- anonymous

on what yu did

- anonymous

umm so basically what i did is that i found that -2 x -10 is equal to positive 20 and add to -12
(negative)x(negative) = positive

- anonymous

oh ok :P

- anonymous

thanks :D

- anonymous

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

- anonymous

or it doesnt matter wht way yu put it as

- anonymous

it doesnt really matter as long as you can get the equation you started with
i believe it looks better for communication the way you put it

- anonymous

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

- anonymous

ok for this one you want a number that adds up to 3 and multiplies to get -18
so you're gonna need a negative and a positive number

- anonymous

any ideas of multiples

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

WOOOOO:P

- anonymous

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

- anonymous

or jus leave it like tht :P

- anonymous

lol i believe you can leave it like that :D

- anonymous

aha ok :P

- anonymous

\[x^{2}-18\]

- anonymous

shud i leave the middle number as 0

- anonymous

what do you think this one is

- anonymous

how wud yu do this
i forgot :$

- anonymous

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

- anonymous

(x-6) (x+6)?

- anonymous

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

- anonymous

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

- anonymous

yu cant?

- anonymous

so yu dun do nytiin to it?

- anonymous

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

- anonymous

good job dada

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