Josh's roots

Note that when $x$ is replaced with -1, the polynomial is equal to 0. Thus -1 is a root. This means $x+1$ is a factor of the polynomial.

Dividing the polynomial by $x+1$ , we get $x^{2}-4x+16$ . Since the two other roots cannot be equal to -1, $w$ and $z$ both satisfy the equation $x^{2}-4x+16=0$

Multiplying the equation by $x+4$ , we have a nicer-looking equation: $x^3+64=0$ Thus the cube of each root of $x^{2}-4x+16$ is -64. This means that $w^{6}=z^{6}=(w^{3})^{2}=(-64)^{2}=(2^{6})^{2}=2^{12}$ which implies that $log_{2}(w^{6}+z^{6})=log_{2}2w^{6}=log_{2}2(2^{12})=log_{2}2^{13}=13$ Therefore, the answer is -1+13=12.

By looking at the polynomial, we see that $x = -1$ is a factor. So, by synthetic division, we get the other factors as:

$(x + 1)(x^{2} - 4x + 16)$

So, $\boxed{r = -1}$

Given, $w$ and $z$ are the roots of $x^{2} - 4x + 16$

Also, $w$ and $z$ are conjugates of each other, as complex roots always occur in conjugate pairs.

Solving the quadratic equation, we get the roots as:

$w$ , $z = 2 \pm i.2 \sqrt{3}$

In polar co-ordinates, it is

$w$ , $z = 4 (cos(60^{o}) \pm i.sin(60^{o}))$

$w^{6}$ , $z^{6} = 4^{6} (cos(60^{o}) \pm i.sin(60^{o}))^{6}$

By De-Moivre's theorem,

$w^{6}$ , $z^{6} = 4^{6} (cos(360^{o}) \pm i.sin(360^{o}))$

$w^{6}$ , $z^{6} = 4^{6}$

$w^{6} + z^{6} = 2.4^{6}$

$w^{6} + z^{6} = 2.2^{12}$

$w^{6} + z^{6} = 2^{13}$

$log_2(w^{6} + z^{6}) = 13$

$r + log_2(w^{6} + z^{6}) = -1 + 13 = 12$

That's the answer!

the polynomial can be factored as: $(x^2-4x+16)(x+1)$ ,therefore $r=-1$ .From vietas formulas we have $r+w+z=3$ $(1)$

$rwz=-16$ $(2)$ where we obtain $w+z=4$ and $wz=16$ .Observe that

$w^6+z^6=(w^2+z^2)(w^4-w^2z^2+z^4)$

$w^2+z^2=(w+z)^2-2wz=-16$ .

$w^4+z^4=(w^2+z^2)^2-2w^2z^2=-256$ and $w^2z^2=256$ ,therefore $w^4-w^2z^2+z^4=-512$ Thus $w^6+z^6=(-16)(-512)=2^{13}$ .Therefore the $\log_2(w^6+z^6)=13$ and the whole expression is equal to $-1+13=\boxed{12}$

Noticing $-1$ is a root of the polynomial, because $1 + 12 = -3 + 16$ , we can factor it into $(x-1)(x^2-4x+16)$ .

Finding the quadratic's roots, we have $w = \frac{4 + \sqrt{4^2-4 \times 16}}{2} \Rightarrow w = 2 + 2i \sqrt{3} \;$ , where $\; i$ is the imaginary unit.

Putting the complex roots into their polar form, we find $w = \sqrt{2^2 + 2 \sqrt{3}^2}(\frac{1}{2} \pm \frac{\sqrt{3}}{2} i ) \Rightarrow w = 4 \; cis (\frac{\pi}{3}) \wedge z = 4 \; cis (\frac{5\pi}{3})$ .

By the De Moivre's Formula, $w^6 = 4^6 \; cis (6 \times \frac{\pi}{3}) \Rightarrow w^6 = 4^6 \; cis (2\pi) \Rightarrow w^6 = 4^6$ and similarly $z^6 = 4^6 \; cis (6 \times \frac{5 \pi}{3}) \Rightarrow z^6 = 4^6 \; cis (10\pi) \Rightarrow z^6 = 4^6$ .

Therefore, $r + log_2 (w^6 + z^6) = -1 + log_2 (2 \times 4^6) = \boxed{12.}$ .

We note that $x=-1$ is a solution to the polynomial. We then use synthetic division to find $x = 2 \pm 2\sqrt{3}i = 4e^{\pm \frac{\pi}{3} i}$ are the other two roots. Then, $w^6 + z^6 = 2^{12} e^{-2\pi} + 2^{12} e^{-2\pi} = 2^{13}$ . Hence, $r + \log_2\left(w^6 + z^6\right) = -1 + 13 = \boxed{12}$

It's pretty easy to see that we can factor out an $x+1$ , leaving the other two roots to be $2(1\pm i\sqrt{3})$ . Now, if we want to compute $w^6 + z^6$ , we could just do it all algebraically, but it should also be interesting to note that all recurrences satisfying $x_{k+3} = 3x_{k+2} - 12x_{k+1} - 16x_k$ have the form $x_k = \alpha r^k + \beta w^k + \gamma z^k$ , where these parameters are solely determined by the initial values, so in order for $\alpha = 0, \beta = \gamma = 1$ , we would need that $x_0 = 1 + 1, x_1 = w + z, x_2 = w^2 + z^2$ . Since $w$ and $z$ are conjugates, $w+z = 2\mathcal{Re}(w)$ and $w^2+z^2 = 2(\mathcal{Re}(w)^2 - \mathcal{Im}(w)^2)$ . Computing these values for higher powers is a chore, but these two values are relatively pleasant to discover. Therefore, we know that $x_0 = 2, x_1 = 4, x_2 = -16, x_3 = -128, x_4 = -256, x_5 = 1024, x_6 = 8192$ , and furthermore $x_k = w^k + z^k$ , so $w^6 + z^6 = 8192$ , which means that the answer is just $13 - 1 = \boxed{12}$ .

By trial and error, we see that $-1$ is a root so $x+1$ is a factor of $x^3 - 3x^2 +12x + 16$ . The toehr factor is $x^2 - 4x+16$ .

$w^6 + z^6 = (w^2+z^2)(w^4 +z^4- (wz)^2)$

$= ((w+z)^2 - 2wz)((w^2 + z^2 )^2 - 3(wz)^2)$

$= (w+z)^2 - 2wz)(((w + z)^2-2wz )^2 - 3(wz)^2)$

$= (4^2 - 2*16)((4^2 - 2(16))^2 - 3*16^2)$

$=-16(16^2 - 3*16^2) = 16(2*16^2) = 2*16^3$

$= 2*(2^4)^3 = 2*2^{12} = 2^{13}$

We have $-1 + log_2 2^{13} = -1+13 = \boxed{12}$ .

(x^3 - 3 x^2 + 12 x + 16) = (x + 1)(x^2 - 4 x + 16) = 0

x = - 1 or 2^2 e^(j Pi/ 3) or 2^2 e^(- j Pi/ 3)

Sum = - 1 + log2 2^12 [e^(j 2 Pi) + e^(- j 2 Pi)] = - 1 + 13 = 12

x^3 − 3x^2+12x+16 = 0 <=> (x+1)(x^2 - 4x +16) = 0 <=> x = -1 => r = -1. And w and z are the two complex roots of x^2 - 4x +16. According to theorem Viète, we have w + z = -b/a = -(-4)/1 = 4, w * z = c/a = 16/1 = 16. w^6 + z^6 = (w + z)^2 * [(w + z)^2 - 3 * w * z]^2 - 2 (w z)^3 = 4^2 - (4^2 - 3*16)^2 - 2 * 16^3 = 8182 = 2^13. So, r + log2(w^6 + r^6) = -1 + log2 (2^13) = 12

-1 is one root of this polynomial. Divide this polynomial by x+1 and get roots of the quadritic polynomial formed and they will be -4omega and -4omega^2. Put these values in the given expression and you will recieve the answer which is 12.

By inspection x=-1 is a root of the given cubic equation. given eqn. can be factorised into (x+1)(x^2-4x+16)

the quadratic has root 2+2* 3^(1/2) i and 2-2 *3^(1/2) i

let w be the 1st one and z be the second img root.

w^2 and z^2 are

8 (3^(1/2) i-1) and 8 (1-3^(1/2) i)
cube both of them to get 8 512 each of them. w^6+z^6=2 8*512=2^13
therefore -1+13=12

Hey Guys! Now again, this happens to be one of Brilliant's highly over-rated problems. So the first step, forwards, is to find the roots of the given polynomial. Trial and error(in this case, it's the first trial itself :P, lucky-lucky!) gives us the real root as -1. Thus,

r = -1

Now dividing the given polynomial by (x + 1), we get a new polynomial--

$x^{2} - 4x + 16$

Solving which, we get the complex roots of the former polynomial as $2 + i(2\sqrt{3})$ and $2 - i(2\sqrt{3})$ . Now, with a bit of manipulation, the two roots can be expressed as

$4(\frac{1}{2} + i(\frac{(\sqrt{3})}{2}$ )) and

$4(\frac{1}{2} - i(\frac{(\sqrt{3})}{2}$ ))

Then, the complex roots can now be expressed as

w = 4cis(60) and

z = 4cis(-60).

$log_{2}$ $(w^{6} + z^{6})$ = $log_{2}$ $(4^{6} + 4^{6})$

which finally equals 13!

Thus, the required expression is -1 + 13 = 12.

Thus the answer!

Cheers!