# factorising hard function

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#1
So this question, the mark scheme factorises to show that f'(x) is always greater than 0 (but that's irrelevant). What I'm trying to figure out is how to factorise such an expression in the first place. I understand that the brackets multiply out to become f'(x), but if I were to come across such a function in an exam I would be at a loss on how to even begin to factorise it into said bracket. It just... doesn't look like it should factorise. Does anyone have any insight or train of thought that can help me understand this more clearly / make it easier to see how to factorise this function, because from my POV the mark scheme basically just pulls it out of thin air.
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1 year ago
#2
Try and factorise 1/x^2 from the first expression, then factorise that quartic, then multiply back in the 1/x^2
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#3
(Original post by Mrepic Foulger)
Try and factorise 1/x^2 from the first expression, then factorise that quartic, then multiply back in the 1/x^2
never mind I forgot to divide 2 by x^-2 aswell brb
Last edited by username3970986; 1 year ago
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#4
(Original post by Mrepic Foulger)
Try and factorise 1/x^2 from the first expression, then factorise that quartic, then multiply back in the 1/x^2
I get stuck at this part at the end where you multiply back in the 1/x^2
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1 year ago
#5
So when you factorise out x^-2 you get x^-2(x^4-2x^2+1) which factorise to x^-2(x^2-1)(x^2-1). Now multiply x^-1 into each of the brackets, and remove the x^-2 in front, giving (x-x^-1)^2
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#6
(Original post by Mrepic Foulger)
So when you factorise out x^-2 you get x^-2(x^4-2x^2+1) which factorise to x^-2(x^2-1)(x^2-1). Now multiply x^-1 into each of the brackets, and remove the x^-2 in front, giving (x-x^-1)^2
ohhhhh, if you multiply 2 brackets by x^-2 each one is multiplied by x^-2 /2 right?
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1 year ago
#7
(Original post by pondering-soul)
ohhhhh, if you multiply 2 brackets by x^-2 each one is multiplied by x^-2 /2 right?
Yes exactly
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#8
(Original post by Mrepic Foulger)
Yes exactly
thank you bro if I ever catch u lackin in the streets imma give u a big sloppy kiss
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1 year ago
#9
(Original post by pondering-soul)
So this question, the mark scheme factorises to show that f'(x) is always greater than 0 (but that's irrelevant). What I'm trying to figure out is how to factorise such an expression in the first place. I understand that the brackets multiply out to become f'(x), but if I were to come across such a function in an exam I would be at a loss on how to even begin to factorise it into said bracket. It just... doesn't look like it should factorise. Does anyone have any insight or train of thought that can help me understand this more clearly / make it easier to see how to factorise this function, because from my POV the mark scheme basically just pulls it out of thin air.
I started by changing 1/x^2 to x^-2 and then you can see you can factorise like a quadratic - effectively if you consider a quadratic to be ax^2 +bx +c, ax^2 = x^2; bx = 2 and c= x^-2 as 2 is effectively 2x^0.

So to factorise x^2 -2x^0 + x^-2 I formed 2 brackets with (x+?)(x+?). Then I dealt with the x^-2 term as the second part of each bracket giving (x-x^-1)(x-x^-1). Then I checked that the middle term could be found from these 2 brackets
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1 year ago
#10
(Original post by pondering-soul)
thank you bro if I ever catch u lackin in the streets imma give u a big sloppy kiss
My guy <3
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