Could someone please help me simplify 3-2(3x-4)(3x+4)-(x+2)(2x-1)?

- Anguyennn

Could someone please help me simplify 3-2(3x-4)(3x+4)-(x+2)(2x-1)?

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- anonymous

Do you know the foil rule for expanding brackets?

- Anguyennn

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

- anonymous

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

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## More answers

- anonymous

Simplify the inside of the brackets firstb to make it easier

- Anguyennn

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

- Anguyennn

oh

- Anguyennn

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

- anonymous

No you can't do that, the -2 is multiplies by brackets its a different type of term than the 3 by itslef, so you cant add them

- anonymous

yeah what you did was right

- Anguyennn

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

- anonymous

but remeber only add things if they have the same thing, for example x^3 and x^2 cant add together, but x^2 and 2x^2 can add to 3x^3

- anonymous

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

- Anguyennn

but what about the 3?

- Anguyennn

on the outside?

- anonymous

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

- Anguyennn

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

- anonymous

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

- Anguyennn

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

- anonymous

oh yeah true that i forgot :P

- Anguyennn

haha its ok

- anonymous

So you have 3-2(9x^2+12x-12x+16)-(2x^2-x+4x-2)
this becomes 3 -2(9x^2 + 16) -(2x^2+3x-2)
this becomes 3 -(18x^2-32) -2x^2-3x+2
this becomes 3 -18x^2 + 32 -2x^2 -3x +2

- anonymous

Now combine like terms

- anonymous

i make a mistake

- anonymous

it should be -32

- anonymous

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

- Anguyennn

no its positive

- Anguyennn

its +32

- anonymous

-2 x 16 = -32

- Anguyennn

-2x-16=32

- anonymous

of i see what was wrong

- Anguyennn

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

- anonymous

i accidentaly put +16 when it was -16

- Anguyennn

and when you multiply them together its -16

- anonymous

yeah

- Anguyennn

multiplied with -2 is 32

- anonymous

sorry for the mistake :P

- Anguyennn

am i correct? haha

- Anguyennn

its ok!!! :)

- anonymous

yeah ur right

- Anguyennn

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

- anonymous

Yep but should the x one be -3x?

- anonymous

yeah its -3x

- Anguyennn

how?

- anonymous

other than that your right, good job

- anonymous

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

- anonymous

in front of the brackets

- Anguyennn

im confused

- Anguyennn

I got -x+4x

- Anguyennn

which gives you 3

- anonymous

You have -(x+2)(2x-1)
this becomes -(2x^2-x+4x-2)
multiplying everyhting by -1
it becomes -2x^2 +x -4x +2
=-2x^2-3x+2

- Anguyennn

why did you multiply everything by -1

- anonymous

when you have brackets and a number outside like a(b+x) , to expand you multiply b and x by a. Outside those brackets theres a -, this just means -1*(2x^2-x+4x-2), therefore everything is multiplied by -1.

- Anguyennn

ohhhh ok i get it now

- anonymous

Ok cool

- Anguyennn

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

- anonymous

yep :)

- Anguyennn

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

- anonymous

Ok if you want

- Anguyennn

so now its factoring polynomial expressions and it asks to factor by finding the greatest common factor. and the expression is 12x^2y^2-18x^3y+24x^2y

- Anguyennn

Thank you for your help

- Anguyennn

I appreciate it

- anonymous

Ok now factoring is just the reverse of the expanding thing I showed you. Expanding is when you have a number multiplied against brackets and then bringing it all out into a polynomial. Factoring is when you have a plynomial and you turn it into a brackets multiplied against a number.

- anonymous

You're very welcome btw, im glad to help

- Anguyennn

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

- anonymous

Ok

- anonymous

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

- anonymous

all the terms have x^2 right?

- anonymous

the middle one is just X^2 x X

- anonymous

to get X^3

- anonymous

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

- Anguyennn

this middle one is different

- anonymous

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

- Anguyennn

yes

- anonymous

Ok those are the common factors, you have to find the highest multiple of every variable (in this case x and y) and the highest multiple of the coefficents

- anonymous

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

- anonymous

Does that make sense?

- Anguyennn

just a little bit

- anonymous

Ok you will get it better in the next step.

- Anguyennn

where did you get 6 from?

- anonymous

Now that we have the highest common factor, we have to find what these multiply against to get each term.

- Anguyennn

oh 6 is the common factor

- anonymous

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

- Anguyennn

yes

- anonymous

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

- anonymous

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

- anonymous

?

- Anguyennn

wait what

- Anguyennn

12?

- anonymous

6(x^2)(y) is the common factor (highest), this means that it can times against a number to make each of the terms

- anonymous

6 x 2 makes 12

- anonymous

so the coeffcient must be 2

- anonymous

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

- anonymous

y x (y ) = y^2

- Anguyennn

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

- Anguyennn

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

- anonymous

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

- anonymous

we are taking each term 1 by 1

- anonymous

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

- Anguyennn

yes please

- anonymous

Ok lets say we have ab + ac

- anonymous

We want to turn it into a bracket times a number

- Anguyennn

ok

- anonymous

the highest common factor in each term is a

- Anguyennn

correct

- anonymous

That goes outside the bracket

- anonymous

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

- anonymous

these go inside the brackets

- Anguyennn

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

- anonymous

just a(b+c)

- anonymous

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

- anonymous

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

- anonymous

Makes sense?

- Anguyennn

yes

- anonymous

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

- anonymous

lets say 2ab+a^2

- anonymous

give it a go

- anonymous

u there?

- Anguyennn

hold on

- Anguyennn

i am just looking it over

- anonymous

k

- anonymous

Remember highest common factor outside brackets, inside goes what it multiplies against to make the orignal terms

- Anguyennn

would it be a?

- anonymous

yep

- Anguyennn

what would the expression look like though?

- anonymous

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

- anonymous

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

- Anguyennn

okk!!!

- Anguyennn

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

- anonymous

Yep :)

- anonymous

you also divide by x^2 and y

- Anguyennn

and then what is the next step?

- anonymous

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

- anonymous

You get 2y when yo do that

- anonymous

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^)

- anonymous

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

- anonymous

do the same for the other 2 terms

- Anguyennn

where did the squared go for the y?

- Anguyennn

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

- anonymous

(y^2)/y = y when you divide variables you minus the power of the bottom one from the power of the top one (aslong they are the same variable)

- anonymous

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

- Anguyennn

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

- anonymous

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

- anonymous

which is just 1

- anonymous

you divide all 3 terms by 6(x^2)y which is the highest common factor to get the number the highest common factor times agasint to get those terms

- Anguyennn

the expression in the middle in x^3 but the other two is x^2 so x^2 is the common factor because there are more of them?

- anonymous

X^3 is just x*x*x X^2 is x*x, the highest is x*x becasue x*x*x is not a factor of the ones with x*x, its bigger than it. x*x can fit into all of them

- anonymous

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

- anonymous

Look at that

- Anguyennn

hold on i am looking at the website

- Anguyennn

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

- anonymous

Yes that is right

- anonymous

Now you got it!

- anonymous

Just do some practice and you'll be sweet.

- Anguyennn

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

- anonymous

Its has to be 6x^2y(2y-3x+12) because remember you turned it into a number times brackets. You can check your answer by expanding, if you get the original expression then its right.

- Anguyennn

oh yes it is

- Anguyennn

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

- Anguyennn

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

- anonymous

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

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