xXnukerrrXx
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Question 18 of Exercise 4D Year 12 Cambridge Can anyone do it? 
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nightweaver066
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18a)
You can write it out in the form of division transformation.
 = (2x+1)(x-3)Q(x) + 3x - 1)
 \textup{ (by remainder theorem)})
b)
Once again, put it in division transformation form.
 = (x^2-1)Q(x) + 2x - 7)
 = 2 - 7)
 = -2 - 7)
Now using the original equation,
 = x^5 + 3x^3 + ax + b)
 = -5 = 1 + 3 + a + b)
 = -9 = -1 -3 -a + b)
Adding these two equations together,
Therefore
c) = (x^2-5)Q(x) + x + 4)
 = (x^2 - 5)Q(-x) -x + 4)
Add these together,
 + P(-x) = (x^2-5)(Q(x) + Q(-x)) + 8)
So,
Should be that, if it's not i'm not too sure.
You can write it out in the form of division transformation.
b)
Once again, put it in division transformation form.
Now using the original equation,
Adding these two equations together,
Therefore
c)
Add these together,
So,
Should be that, if it's not i'm not too sure.
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xXnukerrrXx
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Answer:
18a) 1/2 b)a=-2 and b=-7 c) 8
18a) 1/2 b)a=-2 and b=-7 c) 8
nightweaver066
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b) and c) are correct. a) i'm a little stumped on.
SpiralFlex
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The answer should indeed be 
xXnukerrrXx
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yeh thats what i got before