factorise cubic f without remainder theorm Watch

arslan
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#1
Report Thread starter 13 years ago
#1
3+5x+x²−x³=0

i would use remainder theorm to factorise (and then solve) this cubic function, but the working provided gives the answer in 2 steps:


3+5x+x²−x³=0
(3−x)(1+2x+x²)=0
(3−x)(1+x)²=0
=> x=3, x=-1

how was it done? or did they just use remainder theorm and not show the working. (this is C2 work)

:tsr2:
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thefish_uk
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#2
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(Original post by arslan)
3+5x+x²−x³=0

i would use remainder theorm to factorise (and then solve) this cubic function, but the working provided gives the answer in 2 steps:


3+5x+x²−x³=0
(3−x)(1+2x+x²)=0
(3−x)(1+x)²=0
=> x=3, x=-1

how was it done? or did they just use remainder theorm and not show the working. (this is C2 work)

:tsr2:
Probably by inspection! So basically scratching your head.

A starting point is to notice that the constant, +3, is a prime number so that one factor in the next stage has a 3 in it and one has a 1 in it.
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yusufu
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(Original post by thefish_uk)
Probably by inspection! So basically scratching your head.
You gotta love this emoticon: :hmmmm:

And to the OP, I'd also say they did it by inspection.
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Malik
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#4
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Out of intrest how do you pronounce your name 'arslan'?
Regarding the question I say inspection as well.
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7589200
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#5
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Somebody tipped them off - (3-x) was a factor.
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