# factorising hard function

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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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#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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(Original post by

Try and factorise 1/x^2 from the first expression, then factorise that quartic, then multiply back in the 1/x^2

**Mrepic Foulger**)Try and factorise 1/x^2 from the first expression, then factorise that quartic, then multiply back in the 1/x^2

Last edited by username3970986; 1 year ago

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**Mrepic Foulger**)

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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#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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(Original post by

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

**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

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#7

(Original post by

ohhhhh, if you multiply 2 brackets by x^-2 each one is multiplied by x^-2 /2 right?

**pondering-soul**)ohhhhh, if you multiply 2 brackets by x^-2 each one is multiplied by x^-2 /2 right?

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(Original post by

Yes exactly

**Mrepic Foulger**)Yes exactly

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#9

(Original post by

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.

**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.

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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#10

(Original post by

thank you bro if I ever catch u lackin in the streets imma give u a big sloppy kiss

**pondering-soul**)thank you bro if I ever catch u lackin in the streets imma give u a big sloppy kiss

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