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    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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    (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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    (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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    Out of intrest how do you pronounce your name 'arslan'?
    Regarding the question I say inspection as well.
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    Somebody tipped them off - (3-x) was a factor.
 
 
 
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