The Student Room Group

Factorsing!

2cx^2 +c (6a+b)x+3abc

I know its pretty basic but I am having issues factorising this and its getting on my nerves!!

The closest I get is:

2cx(x+3a) + b(x+3a)

I know it can be further simplified because that isnt the final answer and I am just not confident on the final steps - any help will be really appreciated.
Original post by christinajane
2cx^2 +c (6a+b)x+3abc

I know its pretty basic but I am having issues factorising this and its getting on my nerves!!

The closest I get is:

2cx(x+3a) + b(x+3a)

I know it can be further simplified because that isnt the final answer and I am just not confident on the final steps - any help will be really appreciated.


The first thing you can do is to take out a factor of c from everything. What is left is

2x2+(6a+b)x+3ab\displaystyle 2x^2 + (6a+b)x + 3ab

which you can factor into linear factors in x.
Original post by Zacken
That's great! Both these terms have a common factor of x+3ax+3a, if it helps, why don't you write u=(x+3a)u = (x+3a), then you have:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}2cx u + bu = u(2cx + b) = (x+3a)(2cx+b)\end{equation*}



Isn't there a problem in OP's first step? Or have I misread it?
Reply 3
Original post by christinajane
2cx^2 +c (6a+b)x+3abc

I know its pretty basic but I am having issues factorising this and its getting on my nerves!!

The closest I get is:

2cx(x+3a) + b(x+3a)

I know it can be further simplified because that isnt the final answer and I am just not confident on the final steps - any help will be really appreciated.


See bolded. :-)

Original post by Gregorius
Isn't there a problem in OP's first step? Or have I misread it?


Ah, yes. I assumed that the OP's first step was correct since it looked like the kind of method the question wanted. Having abc at the end makes it rather complicated for what I suspect the OP's level is at, perhaps a typo in the question?
Original post by Zacken

Ah, yes. I assumed that the OP's first step was correct since it looked like the kind of method the question wanted. Having abc at the end makes it rather complicated for what I suspect the OP's level is at, perhaps a typo in the question?


I don't think so; it comes out nice...
Reply 5
Original post by Gregorius
I don't think so; it comes out nice...


Ah, okay. I'll delete my post. Thanks!
Reply 6
Was trying this out for homework and managed to get the answer as (2cx bc)(x 3a)... expanded it out and pretty sure I did it right since I got the original answer.

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