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FP1 Roots of polynomials- HELP!!

I’m stuck on part ii). So for i) I substituted x+2 in to get the cubic 2x^3 +12u^2 +27u+25=0. Is this right?
For ii) I thought that I should do sum of pairs of Roots/ product of Roots but I’m not sure. Anyone able to help?C12C7FAA-D3D1-4DC9-9DD6-9040EA936E0D.jpg.jpeg
Original post by Lexie16
I’m stuck on part ii). So for i) I substituted x+2 in to get the cubic 2x^3 +12u^2 +27u+25=0. Is this right?
For ii) I thought that I should do sum of pairs of Roots/ product of Roots but I’m not sure. Anyone able to help?C12C7FAA-D3D1-4DC9-9DD6-9040EA936E0D.jpg.jpeg


you still have an x term...
Reply 2
It’s meant to be a u, I just typed it out wrong sorry
Original post by the bear
you still have an x term...
Original post by Lexie16
I’m stuck on part ii). So for i) I substituted x+2 in to get the cubic 2u^3 +12u^2 +27u+25=0. Is this right?
For ii) I thought that I should do sum of pairs of Roots/ product of Roots but I’m not sure. Anyone able to help?C12C7FAA-D3D1-4DC9-9DD6-9040EA936E0D.jpg.jpeg


Yes your cubic is correct.

For (ii), do αβ+βγ+αγαβγ=αβαβγ\dfrac{\alpha \beta + \beta \gamma + \alpha \gamma}{\alpha \beta \gamma}= \dfrac{\sum \alpha \beta}{\alpha \beta \gamma}
Original post by Lexie16
I’m stuck on part ii). So for i) I substituted x+2 in to get the cubic 2x^3 +12u^2 +27u+25=0. Is this right?
For ii) I thought that I should do sum of pairs of Roots/ product of Roots but I’m not sure. Anyone able to help?


First part is good.

For the second, note that if x=α,β,γx=\alpha, \beta, \gamma are your roots, then so are u=α2,β2,γ2u=\alpha - 2, \beta -2, \gamma -2 after the substitution.

Hence, if we denote these as u1,u2,u3u_1, u_2, u_3, we are simply looking for 1u1+1u2+1u3\dfrac{1}{u_1} + \dfrac{1}{u_2} + \dfrac{1}{u_3} which you can express in terms of u1u2\displaystyle \sum u_1u_2 and u1u2u3u_1 u_2 u_3 hence read off their values from your new cubic.
(edited 6 years ago)
Reply 5
Original post by BobbJo
Yes your cubic is correct.

For (ii), do αβ+βα+αγαβγ=αβαβγ\dfrac{\alpha \beta + \beta \alpha + \alpha \gamma}{\alpha \beta \gamma}= \dfrac{\sum \alpha \beta}{\alpha \beta \gamma}


Thanks! Sorry to be a pain but I’m having a hard time deciphering that into what I’ve literally got to do. Any way you could write it out and post a pic so it makes sense to me (I understand it’s hard to type out maths)
Reply 6
Original post by Lexie16
Thanks! Sorry to be a pain but I’m having a hard time deciphering that into what I’ve literally got to do. Any way you could write it out and post a pic so it makes sense to me (I understand it’s hard to type out maths)

Logged onto student room on my laptop and I can read it! Don’t worry
so do fraction adding of 1/u1 + 1/u2 + 1/u3 and simplify ...
Original post by Lexie16
Thanks! Sorry to be a pain but I’m having a hard time deciphering that into what I’ve literally got to do. Any way you could write it out and post a pic so it makes sense to me (I understand it’s hard to type out maths)


If a cubic ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0 has roots p,q,rp, q, r, then

p+q+r=bap+q+r=-\dfrac{b}{a}
pq+pr+rq=capq+pr+rq=\dfrac{c}{a}
pqr=dapqr=-\dfrac{d}{a}.

This can be derived by expanding (xp)(xq)(xr)(x-p)(x-q)(x-r) and comparing coefficients.
(edited 6 years ago)
Original post by BobbJo
If the roots of a cubic ax3+bx2+cx+d=0ax^3+bx^2+cx+d=0 are a,b,ca, b, c


Erm... might want to rephrase this.
Original post by RDKGames
hence read off their values from your quadratic.


Original post by RDKGames
Erm... might want to rephrase this.

Thanks! You too :tongue:
Original post by BobbJo
Thanks! You too :tongue:


Long morning :ninja:

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