Factorisation without expansion?

Hi,

I need to factorise the following:-

$8a^2b^3(c^2+1)^4-6a^3b^2(c^2+1)^3+14a^4b(c^2+1)^2$

Can I do this without multiplying out the brackets?

Thanks


Posted from TSR Mobile

Hi,

I need to factorise the following:-

$8a^2b^3(c^2+1)^4-6a^3b^2(c^2+1)^3+14a^4b(c^2+1)^2$

Can I do this without multiplying out the brackets?

Thanks


Posted from TSR Mobile

I doubt you could factorise it if you did expand the brackets.
The trick here is to spot common factors present in all the terms and pull these outside.
Example:

You can apply a similar trick to your question.

If you take out all common factors you should be left with $4b^2(c^2+1)^2-3ab(c^2+1)+7a^2$
Then try putting $X \ in place of \ b(c^2+1)$ and see if you can go on from there.

I am assuming you did the following:-



But what did you cancel next to lead to $4b^2(c^2+1)^2-3ab(c^2+1)+7a^2$?

Thanks and sorry for the silly question.

Rather than dividing, you factor out, so:
$8a^2b^3(c^2+1)^4-6a^3b^2(c^2+1)^3+14a^4b(c^2+1)^2[br]=2a^2b(4b^2(c^2+1)^4-3ab(c^2+1)^3+7a^2(c^2+1)^2)$
Then the next common factor has to be $(c^2+1)^2$ as it is a multiple of each of the terms.
Which when factored out leaves the result you express in the other brackets.

...

So the common factor has to be $(c^2+1)^2$ ?

For some reason when I attempt to do that I end up with the incorrect answer:-

$[(c^2+1)^2]2a^2b(4b^2(c^2+1)^2-3ab(c^2+1)+7a^2$

Obviously my answer is incorrect as I don't see how this could lead to the correct one of $4b^2(c^2+1)^2-3ab(c^2+1)+7a^2$

...

Hi all, I was wondering if any maths buffs out there can help me? I have to sit a maths equivalency test this week, however some answers to the practice paper don't make sense!!

Two questions on factorising algebra are.

1) Factor 6x+12y.....According to the answer sheet the answer is = 6y(y+2)

2) Factorise 4k+8m+24...Apparently the answer should be = (k10) (k+3)

I just cant understand how these answers could be right. What happened to he 'X' in the first equation. Also what happed to the 'M' in the second. Its really confusing me and I can understand how the examiner came to these answers. Help please!!

I am assuming you did the following:-



But what did you cancel next to lead to $4b^2(c^2+1)^2-3ab(c^2+1)+7a^2$?

Thanks and sorry for the silly question.

Hi,

I need to factorise the following:-

$8a^2b^3(c^2+1)^4-6a^3b^2(c^2+1)^3+14a^4b(c^2+1)^2$

Can I do this without multiplying out the brackets?

Thanks

What is the actual question?

If the question is just to factorise -

Your factorisation in post 7 is correct - the answer that you give in that post is incorrect

I didn't cancel anything. I was merely giving what is left after taking out all the commonb factors. i.e. $2a^2b(c^2+1)^2$

Basically the question is to reduce a fraction but it's the factorisation I was struggling with.
So you're saying that $[(c^2+1)^2]2a^2b(4b^2(c^2+1)^2-3ab(c^2+1)+7a^2 \not=4b^2(c^2+1)^2-3ab(c^2+1)+7a^2$?




But how do you do this?

I attempted to factor out the common factor and I still can't see how you got there.....

Basically the question is to reduce a fraction but it's the factorisation I was struggling with.

What I am saying is that
$8a^2b^3(c^2+1)^4-6a^3b^2(c^2+1)^3+14a^4b(c^2+1)^2$
$=2a^2b(c^2+1)^2[4b^2(c^2+1)^2-3ab(c^2+1)+7a^2]$
$=2a^2b(c^2+1)^2[4b(c^2+1)-7a][b(c^2+1)+a]$

Perhaps if you gave us the question we could be more effective in our helping

It's the third line that I am struggling with specifically how we got rid of $7a^2$, $-3ab$ and where $-7a$ came from?