factorise cubic f without remainder theormWatch

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#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)

0
13 years ago
#2
(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)

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.
0
13 years ago
#3
(Original post by thefish_uk)
You gotta love this emoticon:

And to the OP, I'd also say they did it by inspection.
0
13 years ago
#4
Out of intrest how do you pronounce your name 'arslan'?
Regarding the question I say inspection as well.
0
13 years ago
#5
Somebody tipped them off - (3-x) was a factor.
0
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