[bronn]

I'm really stuck on how to solve algebraic fraction problems when there is a squared term on the bottom such as on the attached picture.

I have tried multipyling each term like I would with any normal algebraic fraction where the is no common factor but it all gets real complicated.

[notnek]

You need to look for the lowest common denominator of the fractions. Here, the LCD is

Do you understand that part? Next, change all of the fractions so that becomes the denominator of these fractions. Then you can subtract the fractions in the usual way.

Post your working if you're still stuck.

[JBMJBM]

Removed

[notnek]

(Original post by JBMJBM)
Hello.

I have no idea how to do that fancy handwritten font so it might be hard to visualize my equations.

[1] Times all of the numerators by the other two denominators. Times all of the denominators together.

x+2+3x-1-x^2-4x-4-3x+1-x^2-4x-4-x-2
----------------------------------------------
(x+2)(x+2)(3x^2+5x-2)

[2] Collect like terms of the numerator.

-2x^2-8x-8
-------------
(x+2)(x+2)(3x^2+5x-2)

[3] Factorise top and bottom.

-(2x+4)(x+2)
--------------------------
(x+2)(x+2)(3x-1)(x+2)

[4] Cancel out like terms (x+2).

-(2x+4)
-----------------
(x+2)^2(3x-1)

[5] Expand if necessary

Your method would be faster if you used the LCD that I wrote in my last post instead of the denominator that you used: (x+2)(x+2)(3x^2+5x-2).

[JBMJBM]

Removed

[notnek]

(Original post by JBMJBM)
I haven't actually started C3 yet so...

Could you show more steps of your working please?

I think this is GCSE as opposed to C3.

Often the LCD can be found by multiplying the denominators together which is what you did but in this case, appears in two denominators so the LCD is actually . So you can change all of the fractions so that this becomes their denominator (this first line can be skipped with practice):





Doing it this way stops you from creating an unnecessary factor of on the top and bottom.


In this question, your method is only slightly longer than mine but it's useful to do it the way I suggested because you could find a question like:



where doing it your way would lead to a very complicated numerator.

[JBMJBM]

Removed

[usycool1]

(Original post by JBMJBM)
Is this the correct method?



Because both denominators have at least and we can remove them from both sides of the denominator (and by extension, the numerator) getting this as a result:



Then we have to simplify the numerator:



Then factorise:



Then erase like terms:



Thanks for your help.

You've factorised incorrectly. In fact, I don't think can be factorised any further.

[notnek]

(Original post by JBMJBM)

The answer should be left as shown in the quote above.

Sorry I should've created a question where the numerator could be factorised. But I think you get the idea.

[bronn]

(Original post by notnek)

where doing it your way would lead to a very complicated numerator.

This is what i did which is why I'm getting so confused, thanks for the simpler way.

p.s this is definitely c3 the example was taken straight from a c3 text book.

[Snakefingers13]

I didn't learn this skill until FP1, but I haven't started C3 yet.


This was posted from The Student Room's iPhone/iPad App

[OG-Ashley]

(Original post by bronn)
This is what i did which is why I'm getting so confused, thanks for the simpler way.

p.s this is definitely c3 the example was taken straight from a c3 text book.

what exam board you on? Edexcel?