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All sequences must end with {3rd 0, 1,1,1,1}

Clearly we must start with 0.

We can form

0001111

010101111

00101111

001101111

and that's it.

The probabilities of flipping these sequences is

\(\dfrac{1}{128}+\dfrac{1}{256}+\dfrac{2}{512}=\dfrac{1}{64}\)

\(\text{To start let's assume $a=1,~b=e^{i\theta}$}\\ a+b = 1+\cos(\theta)+i \sin(\theta)\\ |a+b| = (1+\cos(\theta))^2 + \sin^2(\theta) = 2 + 2 \cos(\theta) = 1\\ \cos(\theta) = -\dfrac 1 2\\ \theta = \pm \dfrac{2\pi}{3}\)

\(\text{So now we can let $a=Me^{i\phi},~b = Me^{i (\phi\pm 2\pi/3)}$}\\ \text{and from the previous result we know $|a+b|=M$} \)

\(\dfrac a b = \dfrac{Me^{i\phi}}{Me^{i(\phi \pm 2\pi/3)}} = e^{i\pm 2\pi/3}\\ e^{i2\pi/3}+e^{-i 2\pi/3} = 2\cos\left(\dfrac{2\pi}{3}\right) = -1\)

\(\text{quadratic formula}\\ x = \dfrac{2\pm \sqrt{(-2)^2-(4)(1)(2)}}{2(1)} = \dfrac{2\pm \sqrt{-4}}{2} = 1 \pm i\)

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\(p(x) = x^2 + c_1 x + c_0 \Rightarrow r_1 + r_2 = -c_1\\ f(x) = x^2 - a x + f_0\\ g(x) = x^2 - b x+ g_0\\ h(x) = x^2 - c x + h_0\\ y(x) = f(x) + g(x) + h(x) = 3x^2 -(a+b+c)x + (f_0 +g_0+h_0)\)

\(y(x) = x^2 - \dfrac 1 3(a+b+c)x + \dfrac 1 3 (f_0 + g_0 + h_0)\\ \text{This is the same form as $p(x)$ listed above with $c_1=-\dfrac 1 3(a+b+c)\\$and thus the sum of the roots is $\dfrac 1 3 (a+b+c)$}\)

6C3=20

\(\left(\dfrac 1 a + \dfrac 1 b\right)\left(\dfrac 1 b -\dfrac 1 a\right) = \dfrac{1}{b^2}-\dfrac{1}{a^2}\)

\(\text{to minimize this we want the largest possible magnitude of $b\\$and the smallest possible magnitude of $a$}\\ a = -1,~b=3\\ \dfrac{1}{3^2}-\dfrac{1}{(-1)^2} = \dfrac 1 9 - 1= -\dfrac 8 9\)

\(\text{If $p(x)$ isn't monic, then normalize it so it is. $\\$This won't affect the roots or the sum of the coefficients being 0}\\ p(x) = x^2 + b x -(1+b)\\ p(x) = (x-1)(x+1+b) \text{ with roots}\\ x=1,~x=-(1+b) \text{, both of which are real}\)

a) I assume you can plug 10 into both expressions and evaluate them.  Actually, as the denominators are identical.

you only have to evaluate the numerators at x=10.  Do it and show they are equal at x=10.

b) what do you get in the denominator of both expressions when x = -1/2.  Are we allowed to divide by this number?

c) here you have a bit of algebra to do.  You need to convert the right hand side to an equivalent fraction with (2x+1) in the denominator.

Then you will see that the resulting numerators are the same.