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Do you know the foil rule for expanding brackets?

yes so I got 3-2(9x^2+12x-12x+16)-(2x^2-x+4x-2) is that correct?

Yep thats right, now just times the number outside the brackets by everything inside.

Simplify the inside of the brackets firstb to make it easier

do you minus 3-2 first? .....so it would be 1(9x^2+12x-12x+16)-(2x^2-x+4x-2)?

oh

(9x^2+16)-(2x^2+3x-2)

yeah what you did was right

ok now do I multiply the first bracket with -2?

onceyou multiply the bracket by -2 you can start combining like terms (same power of x)

but what about the 3?

on the outside?

leave that to last, you can only combine it with other numbers by itself (not multiplied by an x)

so right now I am getting 3(-18x^2+32)-(2x^2-x+4x-2)

just leave the 3, and you forgot about the x in the first bracket

but it cancels out doesn't it? because 12x-12x is 0

oh yeah true that i forgot :P

haha its ok

Now combine like terms

i make a mistake

it should be -32

on the last line (i screwed up a couple of lines up)

no its positive

its +32

-2 x 16 = -32

-2x-16=32

of i see what was wrong

because at the beginning its (3x-4)(3x+4)

i accidentaly put +16 when it was -16

and when you multiply them together its -16

yeah

multiplied with -2 is 32

sorry for the mistake :P

am i correct? haha

its ok!!! :)

yeah ur right

so would the answer be -20x^2+3x+33

Yep but should the x one be -3x?

yeah its -3x

how?

other than that your right, good job

in last brackets 4x - x = 3x
as theres a negative in front it becomes -3x

in front of the brackets

im confused

I got -x+4x

which gives you 3

why did you multiply everything by -1

ohhhh ok i get it now

Ok cool

so the answer would now be -20x^2-3x+37

yep :)

since you are helping me do you mind helping me with one more math question? please and thank you

Ok if you want

Thank you for your help

I appreciate it

You're very welcome btw, im glad to help

so could you help me go through this question step by step so that i will get a hang of it please

Ok

First you have to find one number that exist in all 3 terms.

all the terms have x^2 right?

the middle one is just X^2 x X

to get X^3

however the highest y power is only 1 two have y^1 and one has y^2

this middle one is different

all the terms have a multiple of 6, 6 x 2 = 12 6 x 3 = 18 6 x 4 = 24

yes

This means that we can multiply (x^2)(y)(6) by different things to get all the terms

Does that make sense?

just a little bit

Ok you will get it better in the next step.

where did you get 6 from?

oh 6 is the common factor

The 6 is the highest factor of the coeffeicents of each term, it multiplies to make 12 18 and 24

yes

Ok (x^2)(y)(6) * (a+b+c) = the original expression you gave me

what does (x^2)(y)6 times against to make 12x^2y^2 ?

wait what

12?

6 x 2 makes 12

so the coeffcient must be 2

x^2 x 1 = x^2 , we dont worry about this one

y x (y ) = y^2

are you looking at 12x^2y^2 only?

I am confused to which section of the expression you are explaining to me

So the number that 6y(x^2) multiplies agasint to make 12x^2y^2 is 2y

we are taking each term 1 by 1

This seems a bit tedious, ill make an easier example so you understand whats going on ok?

yes please

Ok lets say we have ab + ac

We want to turn it into a bracket times a number

ok

the highest common factor in each term is a

correct

That goes outside the bracket

Now we find the 2 numbers that a multiplies agasint to make ac and ab

these go inside the brackets

so would it look like 2a(b+c)?

just a(b+c)

no if you expand it you get a x b + a x c

which is ab + ac, what we originally had, this is pretty much what factoring is

Makes sense?

yes

ok cool, lets try a harder one then we'll go back to the other

lets say 2ab+a^2

give it a go

u there?

hold on

i am just looking it over

would it be a?

yep

what would the expression look like though?

a((what a times against to make 2ab) + (what a times against to make a^2))

All you're doing is just taking the common factor out leaving everyhting else behind.

okk!!!

ok so for 12x^2y^2 you are dividing 12 by 6 am i correct?

Yep :)

you also divide by x^2 and y

and then what is the next step?

Like this (12x^2y^2)/(6(x^2)y)

You get 2y when yo do that

That is the number which 6(x^2)y times against to make 12x^2y^2
becasue 6(x^2)y * 2y = 12(x^2)(y^)

(y^2)************

do the same for the other 2 terms

where did the squared go for the y?

isn't it suppose to be 6(x^2y^2) ?

for 12x^2y^2 / 6(x^2)y
you split it into 3 parts
12/6 = 2
(x^2)/(x^2) = 1
(y^2)/y = y
2 x 1 x y = 2y

so how come you didn't do (x^2)/x?

it was (x^2)/(x^2) as (x^2) is the highest common factor

which is just 1

https://www.mathsisfun.com/algebra/factoring.html

Look at that

hold on i am looking at the website

would the whole answer be 2y-3x+12

Yes that is right

Now you got it!

Just do some practice and you'll be sweet.

do you have to put 6x^2y(2y-3x+12) or just 2y-3x+12?

oh yes it is

wow thank you so much!!!!!!!!!!!!!!!!!!!!

thanks for helping me so much
it really helped me, now i understand the concept so much

You are welcome :) As long as I teach you something I'll be happy. Good luck for any test you have.