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Roots of Polynomials (Further Maths Year 1)

I've tried this question but things got way too complicated when I got to (ii) so I'm pretty sure I've been doing it wrong. Please can someone help me on how to start it better?
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Reply 1
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MR1999
21
Part ii) should follow naturally from part i) if you've got the correct expressions. You should have

p=−2(αk+α+kα) p = -2\left( \dfrac{\alpha}{k} + \alpha + k\alpha \right)

q=2(α2k+α2+kα2)q=2\left( \dfrac{\alpha^2}{k}+\alpha^2+ k \alpha^2 \right)

r=−2α3r=-2\alpha^3

Then if you factorise p and q properly it shouldn't be too difficult
(edited 5 years ago)
Original post by 0982nR
I've tried this question but things got way too complicated when I got to (ii) so I'm pretty sure I've been doing it wrong. Please can someone help me on how to start it better?


Try harder than you have been trying
Reply 3
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0982nR
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Original post by MR1999
Part ii) should follow naturally from part i) if you've got the correct expressions. You should have

p=−(αk+α+kα) p = -\left( \dfrac{\alpha}{k} + \alpha + k\alpha \right)

q=α2k+α2+kα2q=\dfrac{\alpha^2}{k}+\alpha^2+k\alpha^2

r=−α3r=-\alpha^3

Then if you factorise p and q properly it shouldn't be too difficult



I got this but they were all over 2 as that is what a is?
Reply 4
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MR1999
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Original post by 0982nR
I got this but they were all over 2 as that is what a is?


Yeah I didn't read the question carefully. Now factorise q and p so that they cancel with r
Reply 5
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0982nR
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Original post by MR1999
Yeah I didn't read the question carefully. Now factorise q and p so that they cancel with r


Thank you I have done it now
Original post by MR1999
Part ii) should follow naturally from part i) if you've got the correct expressions. You should have

2p=−(αk+α+kα) 2p = -\left( \dfrac{\alpha}{k} + \alpha + k\alpha \right)

2q=α2k+α2+kα22q=\dfrac{\alpha^2}{k}+\alpha^2+k\alpha^2

2r=−α32r=-\alpha^3

Then if you factorise p and q properly it shouldn't be too difficult


p,q,rp,q,r should be divided by 2.
Reply 7
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MR1999
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Original post by RDKGames
p,q,rp,q,r should be divided by 2.


Thanks.
Reply 8
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0982nR
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Original post by MR1999
Yeah I didn't read the question carefully. Now factorise q and p so that they cancel with r


sorry I think I was wrong. Should I have multiplied all the equations for p,q and r by -2 rather than divided by them. Because a/k+a+ka=-p/2
so p= -2(a/k+a+ak) doesn't it?
Reply 9
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MR1999
21
Original post by 0982nR
sorry I think I was wrong. Should I have multiplied all the equations for p,q and r by -2 rather than divided by them. Because a/k+a+ka=-p/2
so p= -2(a/k+a+ak) doesn't it?


Yeah, you can rewrite the equation as

x3+px2/2+qx/2+r/2=(x−α)(x−kα)(x−α/k)x^3 +px^2/2+qx/2+r/2=(x-\alpha)(x-k\alpha )(x-\alpha /k)

Then equating coefficients gives what we want
(edited 5 years ago)

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