Factorising
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Original post by zed963
So the above is right,
So if i had this expression : 6y(x+3y)+9(x+3y)^2
Would that be (x+3y) and something which I can't figure out HELP
So if i had this expression : 6y(x+3y)+9(x+3y)^2
Would that be (x+3y) and something which I can't figure out HELP
Relax. Take a deep breathe. Now...
6y(x+3y)+9(x+3y)(x+3y)
(x+3y) - Common Factor.
6y
Now continue...
(edited 12 years ago)
Original post by TenOfThem
Yes
What you have there is
6y(x+3y) + 9(x+3y)(x+3y)
Take out the red and what is left ... that is your other bracket
What you have there is
6y(x+3y) + 9(x+3y)(x+3y)
Take out the red and what is left ... that is your other bracket
9y +9 + x
Original post by zed963
9y +9 + x
I am sorry but I am going to bow out now
You need to ask you teacher because you need someone sat next to you pointing to brackets, crossing things out, asking you questions
There is a major gap in your understanding here and you need someone to help properly
Original post by Math12345
Relax. Take a deep breathe. Now...
6y(x+3y)+9(x+3y)(x+3y)
(x+3y) - Common Factor.
6y(x+3y)+9(x+3y)(x+3y)
Now continue...
6y(x+3y)+9(x+3y)(x+3y)
(x+3y) - Common Factor.
6y
Now continue...
why have u crossed the 9 out
Original post by zed963
why have u crossed the 9 out
Look carefully.
Original post by zed963
9y +9 + x
6y
Remove the common factor I have crossed out and you are left with:
6y+9(x+3y)
6y+9x+27y
There's your second bit.
Original post by zed963
why have u crossed the 9 out
Original post by Math12345
Look carefully.
i thought the exact same thing, the 9 isn't crossed out, just how the 9 is displayed next to something that IS crossed out, makes it look like it is crossed out.
As TenOfThem said, you really need to go and see a teacher, or even ask a friend for some help. With these sort of things, you do need people to point it out to you, otherwise you'll just keep posting question after question.
Original post by Math12345
6y(x+3y)+9(x+3y)(x+3y)
Remove the common factor I have crossed out and you are left with:
6y+9(x+3y)
6y+9x+27y
There's your second bit.
Remove the common factor I have crossed out and you are left with:
6y+9(x+3y)
6y+9x+27y
There's your second bit.
So that would be 3(3x+11y)
Original post by zed963
So that would be 3(3x+11y)
Leave it as (9x+33y).
Original post by zed963
So that would be 3(3x+11y)
But in the answer sheet it give 3(x+y)(3x+11y)
Original post by zed963
But in the answer sheet it give 3(x+y)(3x+11y)
wouldn't matter. Either is correct.
Original post by zed963
But in the answer sheet it give 3(x+y)(3x+11y)
Talk to a teacher, because I don't think you get it completely.
(edited 12 years ago)
Original post by Math12345
Talk to a teacher, because I don't think you get it completely.
but the common factor of 6 and 9 is three
Original post by zed963
but the common factor of 6 and 9 is three
Yes, but I still think you don't know how to get there.
Practice a few more questions.
Original post by Math12345
Yes, but I still think you don't know how to get there.
Practice a few more questions.
Practice a few more questions.
Have you got anymore questions?
Original post by zed963
Have you got anymore questions?
No, I suggest you ask your teacher first for a good detailed explanation then ask them for questions.
Hi, I'm going to hijack this thread cos the topic is pretty similar. I'm self-teaching A-level and I unfortunately think I have missed something basic in factorising skills that keeps bringing me to a dead stop. It happens when the coefficient of x^2 is particularly high, especially when the resulting ac is large and negative. This is the problem that has just confused me at this point:
16x^2 + 16x - 45
I couldn't work out for the life of me any numbers that worked and when I checked the answer is (4x + 9)(4x - 5). I get that there was (probably?) a (16x + 36)(16x - 20) stage but I don't think I could ever have got to it by myself.
How do I go about finding the right figures? It's worst when b is positive and ac is negative, is there a method or a stage to the method I have missed to make it easier to figure out correctly? Thanks a lot for any help.
16x^2 + 16x - 45
I couldn't work out for the life of me any numbers that worked and when I checked the answer is (4x + 9)(4x - 5). I get that there was (probably?) a (16x + 36)(16x - 20) stage but I don't think I could ever have got to it by myself.
How do I go about finding the right figures? It's worst when b is positive and ac is negative, is there a method or a stage to the method I have missed to make it easier to figure out correctly? Thanks a lot for any help.
Original post by Ronove
Hi, I'm going to hijack this thread cos the topic is pretty similar. I'm self-teaching A-level and I unfortunately think I have missed something basic in factorising skills that keeps bringing me to a dead stop. It happens when the coefficient of x^2 is particularly high, especially when the resulting ac is large and negative. This is the problem that has just confused me at this point:
16x^2 + 16x - 45
I couldn't work out for the life of me any numbers that worked and when I checked the answer is (4x + 9)(4x - 5). I get that there was (probably?) a (16x + 36)(16x - 20) stage but I don't think I could ever have got to it by myself.
How do I go about finding the right figures? It's worst when b is positive and ac is negative, is there a method or a stage to the method I have missed to make it easier to figure out correctly? Thanks a lot for any help.
16x^2 + 16x - 45
I couldn't work out for the life of me any numbers that worked and when I checked the answer is (4x + 9)(4x - 5). I get that there was (probably?) a (16x + 36)(16x - 20) stage but I don't think I could ever have got to it by myself.
How do I go about finding the right figures? It's worst when b is positive and ac is negative, is there a method or a stage to the method I have missed to make it easier to figure out correctly? Thanks a lot for any help.
With those, I think you just have to 'see' it. Maybe someone has a better method (EDIT: Turns out Maths12345 does
), but if you can't see it then I'd just complete the square.(edited 12 years ago)
Original post by Ronove
Hi, I'm going to hijack this thread cos the topic is pretty similar. I'm self-teaching A-level and I unfortunately think I have missed something basic in factorising skills that keeps bringing me to a dead stop. It happens when the coefficient of x^2 is particularly high, especially when the resulting ac is large and negative. This is the problem that has just confused me at this point:
16x^2 + 16x - 45
I couldn't work out for the life of me any numbers that worked and when I checked the answer is (4x + 9)(4x - 5). I get that there was (probably?) a (16x + 36)(16x - 20) stage but I don't think I could ever have got to it by myself.
How do I go about finding the right figures? It's worst when b is positive and ac is negative, is there a method or a stage to the method I have missed to make it easier to figure out correctly? Thanks a lot for any help.
16x^2 + 16x - 45
I couldn't work out for the life of me any numbers that worked and when I checked the answer is (4x + 9)(4x - 5). I get that there was (probably?) a (16x + 36)(16x - 20) stage but I don't think I could ever have got to it by myself.
How do I go about finding the right figures? It's worst when b is positive and ac is negative, is there a method or a stage to the method I have missed to make it easier to figure out correctly? Thanks a lot for any help.
First thing to always check is if you can simplify the equation. In this case you can't.
16x^2+16x-45
The product is a*c: 16*-45=-720
The sum is b: +16
What 2 numbers give a product of -720 and a sum of +16?
36 and -20.
Now split the b term into this: (you should put it in a order, so you can factorise e.g. 16 and 36 (have 4 in common) and 20 and 45 (have 5 in common).
16x^2+36x-20x-45
Factorise
4x(4x+9)-5(4x+9)
Common factor?
(4x+9)
Next bracket is the bits left over.
(4x-5)
Answer:
(4x+9)(4x-5)
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