supreme
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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


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joostan
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(Original post by supreme)
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


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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:
Spoiler:
Show
7 \times x^2 \times 9 -9 \times 7 \times x +x \times  64 \times 7
Spot that 7and x are common to both:
Then pull out the 7x to:
 7x(9x-9+64)
Simplify further to: 7x(9x+55)

You can apply a similar trick to your question.
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brianeverit
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(Original post by supreme)
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


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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.
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supreme
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(Original post by joostan)
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:
Spoiler:
Show
7 \times x^2 \times 9 -9 \times 7 \times x +x \times  64 \times 7
Spot that 7and x are common to both:
Then pull out the 7x to:
 7x(9x-9+64)
Simplify further to: 7x(9x+55)

You can apply a similar trick to your question.
Thanks.

So should I attempt to factorise 8a^2b^3-6a^3b^2+14a^4b?
I'm guessing not! It's the fact, it's not expanded that I struggle with.
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supreme
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(Original post by brianeverit)
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:-

\dfrac{8a^2b^3(c^2+1)^4-6a^3b^2(c^2+1)^3+14a^4b(c^2+1)^2}{2a^2b}\left=4b^2(c^2+1)^4-3ab(c^2+1)^3+7a^2(c^2+1)^2

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.
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joostan
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(Original post by supreme)
I am assuming you did the following:-

\dfrac{8a^2b^3(c^2+1)^4-6a^3b^2(c^2+1)^3+14a^4b(c^2+1)^2}{2a^2b}\left=4b^2(c^2+1)^4-3ab(c^2+1)^3+7a^2(c^2+1)^2

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

=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.
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supreme
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(Original post by joostan)
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

=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
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lazy_fish
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(Original post by supreme)
...
Looks correct.
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joostan
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#9
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(Original post by supreme)
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
:confused:
The result is the same. . . you have the correct answer multiplied by a common factor which is sitting out the front.
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LincsLady
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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!!
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lazy_fish
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(Original post by LincsLady)
...
Does not take a 'maths buff' to figure out those questions or answers are wrong.
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LincsLady
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Thanks. I'm terrible at maths but I thought surely a reputable examination company cant print the wrong answers on the answer sheet. Obviously they can!
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Smaug123
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(Original post by LincsLady)
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!!
The first one would be 6y^2+12y to correspond with the given answer. The second (assuming you meant to write (k+10)(k+3)) would be k^2+13k+30.
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brianeverit
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(Original post by supreme)
I am assuming you did the following:-

\dfrac{8a^2b^3(c^2+1)^4-6a^3b^2(c^2+1)^3+14a^4b(c^2+1)^2}{2a^2b}\left=4b^2(c^2+1)^4-3ab(c^2+1)^3+7a^2(c^2+1)^2

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.
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
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TenOfThem
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(Original post by supreme)
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
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supreme
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(Original post by TenOfThem)
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
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?


(Original post by brianeverit)
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
But how do you do this?

I attempted to factor out the common factor and I still can't see how you got there.....
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brianeverit
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(Original post by supreme)
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.....
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]
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TenOfThem
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(Original post by supreme)
Basically the question is to reduce a fraction but it's the factorisation I was struggling with.
Perhaps if you gave us the question we could be more effective in our helping
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supreme
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(Original post by brianeverit)
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]
(Original post by TenOfThem)
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?
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brianeverit
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(Original post by supreme)
It's the third line that I am struggling with specifically how we got rid of 7a^2, -3ab and where -7a came from?
Write b(c^2+1)=X \text{ and }a=Y then we are just factorising 4X^2-3XY+7Y^2
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