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kjyugjhvl

jgfjgjghh

(3x-1)(3x-1)(3x-1)=[(3x-1)(3x-1)](3x-1)

simply ignore the final bracket until you've worked out the ones in the [] brackets, then using the same method mulitply out what you get from [] with the (3x-1) on the end :smile:

EDIT: fatchicksftw.

Reply 2

steve2005

17

(3x-1)(3x-1)(3x-1)=(3x-1)(9x^2-6x+1)

9x^2 -6x + 1

Binomial expansion??

a)Try doing just simply (3x-1)(3x-1) and what do you get?

b)Then you do (3x-1)(your answer from part a)

Reply 6

Totally Tom

18

sorry, did i just MERK YOU ALL!

muhahahaha.

Reply 7

Gaara.

1

steve2005 laid out the way i would explain doing it OP

Reply 8

ilovefatchicks

OP

3

steve2005

(3x-1)(3x-1)(3x-1)=(3x-1)(9x^2-6x+1)

I can get the 9x^2 and the +1 but I cannot work out how you get the 6x when you merge out the two brackets.
Could you please explain

Reply 9

nuodai

17

Expanding the second two brackets is just like expanding a quadratic function

(ax+b)(cx+d)

; then you just multiply through again.

Say you have something with 5 brackets (...)(...)(...)(...)(...), then you can tackle it by multiplying everything in the 4th bracket with everything in the 5th, and then everything in the 3rd with what you just worked out for the 4th and 5th, then everything in the 2nd with everything that you just worked out for the 3rd, 4th and 5th, and so on... for example:

\newline (a+b)(c+d)(e+f)(g+h)(i+j)\newline[br]= (a+b)(c+d)(e+f)(gi + gj + hi + hj)\newline[br]= (a+b)(c+d)(egi + egj + ehi + ehj + fgi + fgj + fhi + fhj)\newline[br]= (a+b)(cegi + cegj + cehi + cehj + cfgi + cfgj + cfhi + cfhj + degi + degj + dehi + dehj + dfgi + dfgj + dfhi + dfhj)\newline[br]= acegi + acegj + acehi + acehj + acfgi + acfgj + acfhi + acfhj + adegi + adegj + adehi + adehj + adfgi + adfgj + adfhi + adfhj + bcegi + bcegj + bcehi + bcehj + bcfgi + bcfgj + bcfhi + bcfhj + bdegi + bdegj + bdehi + bdehj + bdfgi + bdfgj + bdfhi + bdfhj

Simples *squeak*

But really, expanding brackets is quite basic GCSE stuff... what level are you working at here?

Reply 10

qgujxj39

17

ilovefatchicks

I can get the 9x^2 and the +1 but I cannot work out how you get the 6x when you merge out the two brackets.
Could you please explain

(a + b)^2 = (a + b)(a + b) = a(a+b) + b(a+b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2

Apply this to your brackets.

Reply 11

PGtips92

12

OMG a maths question on tsr i can actually do!!!

So,

(3x-1)(3x-1)(3x-1)
= (3x-1)(9x^2 -3x - 3x +1)
= 27x^3 - 9x^2 -9x^2 + 3x
-9x^2 +3x +3x -1
= 27x^3 - 27x^2 + 9x -1

(probably beaten to it but heyho..)

EDIT: why does it say -2 at the end?!? Sorry if i'm wrong!!

Reply 12

Banburyhammer

1

ilovefatchicks

I can get the 9x^2 and the +1 but I cannot work out how you get the 6x when you merge out the two brackets.
Could you please explain

Thats because you multiply 3x by -1 to get -3x, and then again. You then add the answers together like so: -3x + -3x = -6x

Remember to do that as well as 3x*3x and -1*-1.

Reply 13

ste0731

12

(3x-1) * (3x-1)

FOIL
First = 3x * 3x = 9x^2
Outside = -1*-1 = +1
inside = -1 * 3x = -3x
Last = -1 * 3x = -3x
collect terms

9x^2-6x+1

Reply 14

Totally Tom

18

nuodai

Expanding the second two brackets is just like expanding a quadratic function

(ax+b)(cx+d)

; then you just multiply through again.

Say you have something with 5 brackets (...)(...)(...)(...)(...), then you can tackle it by multiplying everything in the 4th bracket with everything in the 5th, and then everything in the 3rd with what you just worked out for the 4th and 5th, then everything in the 2nd with everything that you just worked out for the 3rd, 4th and 5th, and so on... for example: