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Factorising

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Original post by zed963
I am also having trouble with this thing

(x-y)^2-b(x-y)


maybe it would help you to "substitute" (x-y) with another letter, say t?

so you'd have
t^2 - bt

and then factorise that out, as you understood how to do above.




you'd get t(t-b)
and now put back in your substitution for t... t = (x-y)

so you'd have (x-y)((x-y)-b)

the same goes for the (x+5) one:
(x+5)+3(x+5)^2
let x+5 = t

so it becomes
t+3t^2
factorise:
t(1+3t)

put back in t=(x+5)

(x+5)(1+3(x+5)) and simplify
Reply 21
Original post by Math12345
1. Write out the quadratic like this: (x-y)(x-y)-b(x-y)
2. Common factor? (...)
3. Then you collect the other terms in the other bracket (...)


(x-y) But you see thats the thing I am not able to spot that immeadiately is there any resource that can teach you this thing.
Original post by zed963
(x-y) But you see thats the thing I am not able to spot that immeadiately is there any resource that can teach you this thing.


That's why you should write (x-y)^2 as (x-y)(x-y) - It's much easier to spot the common factor then.
Original post by zed963
Now thats the bit where I get confused.


So

y+3y2=y(1+3y)y+3y^2 = y(1+3y)

m+3m2=m(1+3m)m+3m^2 = m(1+3m)

ab+3(ab)2=ab(1+3ab)ab+3(ab)^2 = ab(1+3ab)

(Zed)+3(Zed)2=Zed(1+3Zed)(Zed)+3(Zed)^2 = Zed(1+3Zed)

(anything)+3(anything)2=(anything)(1+3(anything))(anything)+3(anything)^2 = (anything)(1+3(anything))

(x+5)+3(x+5)2=(x+5)(1+3(x+5))(x+5)+3(x+5)^2 = (x+5)(1+3(x+5))
(edited 11 years ago)
Reply 24
Original post by TenOfThem
So

y+3y2=y(1+3y)y+3y^2 = y(1+3y)

m+3m2=m(1+3m)m+3m^2 = m(1+3m)

ab+3(ab)2=ab(1+3ab)ab+3(ab)^2 = ab(1+3ab)

(Zed)+3(Zed)2=Zed(1+3Zed)(Zed)+3(Zed)^2 = Zed(1+3Zed)

(anything)+3(anything)2=(anything)(1+3(anything))(anything)+3(anything)^2 = (anything)(1+3(anything))

(x+5)+3(x+5)2=(x+5)(1+3(x+5))(x+5)+3(x+5)^2 = (x+5)(1+3(x+5))


So if we were to expand this whole equation what would it become: (x+5)+3(x+5)^2
If you expand (x+5)+3(x+5)2=x+5+3x2+30x+75=3x2+31x+80(x+5) + 3(x+5)^2 = x+5 + 3x^2+30x+75 = 3x^2 + 31x + 80

If you meant how do we fully factorise

(x+5)+3(x+5)2=(x+5)(1+3(x+5))=(x+5)(1+3x+15)=(x+5)(3x+16)(x+5) + 3(x+5)^2 = (x+5)(1+3(x+5)) = (x+5)(1+3x+15) = (x+5)(3x+16)
Reply 26
Original post by Math12345
That's why you should write (x-y)^2 as (x-y)(x-y) - It's much easier to spot the common factor then.


So i f i have one common factor then what do i do after?
Reply 27
Original post by TenOfThem
If you expand (x+5)+3(x+5)2=x+5+3x2+30x+75=3x2+31x+80(x+5) + 3(x+5)^2 = x+5 + 3x^2+30x+75 = 3x^2 + 31x + 80

If you meant how do we fully factorise

(x+5)+3(x+5)2=(x+5)(1+3(x+5))=(x+5)(1+3x+15)=(x+5)(3x+16)(x+5) + 3(x+5)^2 = (x+5)(1+3(x+5)) = (x+5)(1+3x+15) = (x+5)(3x+16)


I get the expand bit but I've been taught to find the factors of 240 that make 31 and then split it in half and factorise, thats why i'm not familiar with the common factor thing.
Original post by zed963
I get the expand bit but I've been taught to find the factors of 240 that make 31 and then split it in half and factorise, thats why i'm not familiar with the common factor thing.


Oh ... you mean to factorise the

3x2+31x+803x^2 + 31x + 80

version

well that is fine ... why not just use that
(edited 11 years ago)
Reply 29
Original post by TenOfThem
Oh ... you mean to factorise the

3x2+31x+803x^2 + 31x + 80

version

well that is fine ... why not just use that


But I don't think that works for (x-y)^2 -b(x-y)
Original post by zed963
I get the expand bit but I've been taught to find the factors of 240 that make 31 and then split it in half and factorise, thats why i'm not familiar with the common factor thing.


If they give you the quadratic like that then you should do that, but if they give you a common factor then it is much easier to use that.
(edited 11 years ago)
(x−y)2−b(x−y)=(x−y)(x−y−b)(x-y)^2 - b(x-y) = (x-y)(x-y-b)

is clearly the easiest way

I am not really sure what you are struggling with in the above

Common factor simply means the same thing in both parts and it is clear that there is an (x-y) in both bits
Reply 32
Original post by TenOfThem
(x−y)2−b(x−y)=(x−y)(x−y−b)(x-y)^2 - b(x-y) = (x-y)(x-y-b)

is clearly the easiest way

I am not really sure what you are struggling with in the above

Common factor simply means the same thing in both parts and it is clear that there is an (x-y) in both bits


Yes I get the common factor bit but not how you get (x-y-b)
(x-y)(x-y)-b(x-y)
Common factor: (x-y)

Cross it out above, so you are left with.
(x-y)-b

Put this into one bracket:
(x-y-b)

Therefore: (x-y)(x-y-b)
(edited 11 years ago)
Reply 34
Original post by Math12345
(x-y)(x-y)-b(x-y)
Common factor: (x-y)

Cross it out above, so you are left with.
(x-y)-b

Put this into one bracket:
(x-y-b)

Therefore: (x-y)(x-y-b)


Oh right thanks, Can you give me a similar expression so that I can factorise it?
Original post by zed963
Oh right thanks, Can you give me a similar expression so that I can factorise it?


(x+2y)^2-400b(x+2y)
Reply 36
Original post by Math12345
(x+2y)^2-400b(x+2y)


(x+2Y)(-400bx+2y)
Original post by zed963
(x+2Y)(-400b+x+2y)


I would write it as: (x+2y)(x+2y-400b)
Reply 38
Original post by Math12345
I would write it as: (x+2y)(x+2y-400b)


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
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


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

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