Edexcel Core Pure Maths Book 1/AS quartic equation question 
From the book in the title, this is page 61, Ex.4C, Q9. I found all the other questions in the exercise (4C) relatively straightforward, but this one stumped me. I suppose you could guess a few values of alpha and r, but I'm not sure in an exam I would think I would be expected to "guess" 1/2 or -1/2 as a root.
I've tried a bunch of stuff and have shown my attempts below, but you can see that towards the bottom of my page of working I end up going round in circles and I'm not making any progress towards the answer.
Obviously I'm looking to use the methods they have given in the book to answer the questions, which are the equations immediately below.
Any help will be appreciated.


 A: We need to solve $$1024x^4-960x^3+280x^2-30x+1=0.$$
Now, for all real $k$ we obtain:
$$1024x^4-960x^3+280x^2-30x+1=$$
$$=(32x^2-15x+k)^2-((64k-55)x^2-30(k-1)x+k^2-1),$$
which for $k=1$ gives:
$$1024x^4-960x^3+280x^2-30x+1=$$
$$=(32x^2-15x+1)^2-9x^2=(32x^2-18x+1)(32x^2-12x+1).$$
Can you end it now?
In the first step I used the following formula:
$$(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc.$$
Since $$(32x^4-15x+k)^2=1024x^4+225x^2+k^2-960x^3+64kx^2-30x,$$
we obtain:
$$1024x^4-960x^3+280x^2-30x+1=$$
$$=1024x^4+225x^2+k^2-960x^3+64kx^2-30x-64kx^2+55x^2+30kx-30x-k^2+1=$$
$$=(32x^2-15x+k)^2-((64k-55)x^2-30(k-1)x+k^2-1).$$
A: I had the same thought as Lulu: if you factor out the 3, you get $$1024x^4-960x^3+280x^2-30x+1=0$$  So $\alpha(r\alpha)(r^2\alpha)(r^3\alpha)=r^6\alpha^4=\frac{1}{1024}$ (from your problem hints) or $r^3\alpha^2=\frac{1}{32}$ and $r=\alpha=\frac{1}{2}$ is a really quick first guess.
A: As Dave L. Renfro pointed out in a comment, note that, from the $ \sum \alpha\beta$ formulae, we have:
$\frac{35}{128} = \alpha^{2}(r+r^{2}+2r^{3}+r^{4}+r^{5})= \alpha^{2}r^{3}(\frac{1}{r^{2}}+\frac{1}{r}+2+r+r^{2})=\frac{1}{32}(S^{2}+S)$ where $S = \frac{1}{r}+r$.
Solve for $S$ to get a quadratic in $r$, then solve for $r$ etc.
