Question

Factor the ff. by finding the common minomial factor :

1. 2x-18
2. 3x-12
I NEED ALL YOUR HELP !! PLEASE HELP ME !! :''((

Answer 1

2x-18 = 2x - 2*9 = 2(x-9)
3x - 12 = 3x - 3*4 = 2(x-4)

Answer 2

* replace 2 with 3 in second

Answer 3

hello . what is the meaning of this sign ? is it the exponent ?

Answer 4

Ok... below is the answer.
2x-18 = 2x - 2*9 = 2(x-9)
3x - 12 = 3x - 3*4 = 3(x-4)

"*" is for comment or PS like thing .... lol. I confused you

Answer 5

1. 2 (x-9)
2. 3 (x-4)

Answer 6

who are you queensen ? don't you see ? ms ramya teaching me !!

Answer 7

lol im no good in teaching... :/

Answer 8

thanks you very much ms. ramya

Answer 9

it's alright i need help again

Answer 10

No problem :)

Answer 11

Groupings of \[2x^{2}-x+2xy\]

Answer 12

uhm, mathslover,

Answer 13

\[\huge{x(2x-1+2y)}\]
i took x common from the given expression ..

Answer 14

how do you get that ?

Answer 15

given :
\[\huge{2x^2-x+2xy}\]
take x common :
\[\huge{x(2x-1+2y)}\] Did u get it now ?

Answer 16

no, i need steps

Answer 17

ok :
\[\huge{2x^2-x+2xy}\]
\[\huge{2(x*x)-(1*x)+2(x*y)}\]
\[\huge{2x(x)-1(x)+2y(x)}\]
\[\huge{x(2x-1+2y)}\] got it now ?

Answer 18

yup, but in this equations, factor the ff. by finding the common monomial factor
\[x^{2}-5x\]

Answer 19

whats the answer, and how to get that ?

Answer 20

well, what is the common factor in the equation?
Hint: there are two \(x\)

Answer 21

\[\huge{x^2-5x}\]
\[\huge{(x*x)-(5*x)}\]
\[\huge{x(x-5)}\]

Answer 22

what is that ?

Answer 23

process , i took x common

Answer 24

aa ..

Answer 25

mathslover, do all you required to give the answers or only the hints ?

Answer 26

pls answer

Answer 27

we are just said to give hints .. where r u having problem

Answer 28

aa .. i need the answers because i have so many assignments .. :(

Answer 29

asking for the answers is prolly going to take a long time if you get any at all...
your best bet is to understand what everyone is teaching so you can carry out the process yourself...
when you do the problems yourself, you'll find that this thing really isn't that tough....

there is nothing wrong with asking for help when you get stuck on a particular problem.... we'd be glad to help out in that case.

Answer 30

:))

Answer 31

still need help/explanation? :)

Answer 32

yup

Answer 33

factor the ff. by groupings
abc plus 2c plus 3ab plus 6

Answer 34

\[\large abc+2c+3ab+6 \]
\[\large abc+3ab+2c+6 \]
\[\large (abc+3ab)+(2c+6) \]
\[\large (\color{red}{ab}c+3\color{red}{ab})+(\color{red}{2}c+\color{red}2\cdot 3) \]
\[\large \color{red}{ab}(c+3)+\color{red}2(c+3) \]
\[\large ab\color{blue}{(c+3)}+2\color{blue}{(c+3)} \]
\[\large (ab+2)(c+3) \]