# Polynomial Substitution Method

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#1
Is anyone able to help with the below Further Maths question?

I'm unsure if / where I've gone wrong in part i - but ended up with my signs different to the answer given in my textbook.
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#2
(Original post by beachpanda)
Is anyone able to help with the below Further Maths question?

I'm unsure if / where I've gone wrong in part i - but ended up with my signs different to the answer given in my textbook.
The textbook answer seems to be valid for substitution z = 2 + w

But I think my method was correct by using z = 2 - w as I did the same for other questions and they were fine?
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1 month ago
#3
(Original post by beachpanda)
Is anyone able to help with the below Further Maths question?

I'm unsure if / where I've gone wrong in part i - but ended up with my signs different to the answer given in my textbook.

Sum of roots of original polynomial is -4/2= -2

So, sum of new roots is 6-(-2) = 8

So, new poly. will start
Last edited by ghostwalker; 1 month ago
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1 month ago
#4
I think you're right and the textbook is wrong. (Or at least I agree on the sign of the x^2 coeff, which is all I checked).
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1 month ago
#5
You're correct for i). Must be a book typo. Guess it's similar for ii).
Just chuck them into wolframalpha
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#6
(Original post by ghostwalker)

Sum of roots of original polynomial is -4/2= -2

So, sum of new roots is 6-(-2) = 8

So, new poly. will start
(Original post by DFranklin)
I think you're right and the textbook is wrong. (Or at least I agree on the sign of the x^2 coeff, which is all I checked).
(Original post by mqb2766)
You're correct for i). Must be a book typo. Guess it's similar for ii).
Just chuck them into wolframalpha
I was hoping so, thanks all!
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