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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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)
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
yeah what you did was right
ok now do I multiply the first bracket with -2?
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
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
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
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
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
why did you multiply everything by -1
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.
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
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
Thank you for your help
I appreciate it
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.
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
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
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?
Now that we have the highest common factor, we have to find what these multiply against to get each term.
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)(y) is the common factor (highest), this means that it can times against a number to make each of the terms
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
k
Remember highest common factor outside brackets, inside goes what it multiplies against to make the orignal terms
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) ?
(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)
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
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
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?
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
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?
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.
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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question