MaxWong

The longest line segment is the main diagonal.

If you want the length of it:

$\begin{array}{cl} &\text{Length of main diagonal}\\ =& \sqrt{14^2 + 9^2 + 7^2}\\ =& \sqrt{326} \end{array}$

We have the formula:

$\operatorname{Proj}_{\mathbf{w}} (\mathbf v) = \dfrac{\langle\mathbf v, \mathbf w\rangle}{\|\mathbf w\|^2}\mathbf{w}$

Now, we compute $\langle\mathbf v, \mathbf w\rangle$ and $\|\mathbf w\|^2$.

$\begin{array}{cl} &\langle\mathbf v, \mathbf w\rangle\\ =& 2(2) + 3(-1) + (-1)(0)\\ =& 1\\ &\|\mathbf w\|^2\\ =& 2^2 + (-1)^2 + 0^2\\ =&5 \end{array}$

Therefore, substituting back into the formula, we have

$\operatorname{Proj}_{\mathbf{w}} (\mathbf v) = \dfrac15\mathbf{w} = \begin{pmatrix}\frac25\\-\frac15\\0\end{pmatrix}$

We assume $p(x) = x^4 + ax^3 + bx^2 + cx + d$.

Since p(1) = 3, $a + b + c + d = 2$.

Since p(3) = 11, $27a + 9b + 3c + d = -70$.

Since p(5) = 29, $125a + 25b + 5c + d = -596$.

This system has 3 equations but 4 variables, so there are infinitely many solutions.

Solving, we have $\begin{cases}b = -9a - \dfrac{227}4\\c = 23a + 191\\d = -15a - \dfrac{529}4\end{cases}$.

With these coefficients, we have $p(-2) = 16 - 8a + 4\left(-9a - \dfrac{227}4\right) - 2(23a + 191) + \left(-15a - \dfrac{529}4\right) = -\dfrac{2901}4 - 105 a$

Also, $p(6) = 1296 + 216a + 36\left(-9a - \dfrac{227}4\right) + 6(23a + 191) + \left(-15a -\dfrac{529}4\right) = \dfrac{1067}4 + 15 a$.

Hence, 

$\begin{array}{rcl}p(-2) + 7p(6) &=& -\dfrac{2901}4 - 105a + 7\left(15a + \dfrac{1067}4\right) \\ &=&1142\end{array}$

I wrote a C++ program to simulate the situation.

EDIT: This part was originally the code, but the website removes angle brackets automatically, so I resorted to pastebin for the code.

Note that $150! = 1\times 2 \times 3 \times 4\times \cdots$.

To find the largest n for which 12^n divides 150!, we find the largest m and k for which 2^m divides 150! and 3^k divides 150!.

The formula for finding the largest power n of a prime p for which p^n divides k! is: $\max n = \displaystyle\sum_{r = 1}^\infty \left\lfloor\dfrac{k}{p^r}\right\rfloor$.

Then, $\max m = \displaystyle\sum_{r = 1}^\infty \left\lfloor\dfrac{150}{2^r}\right\rfloor=\left\lfloor\dfrac{150}{2}\right\rfloor+ \left\lfloor\dfrac{150}{2^2}\right\rfloor+ \left\lfloor\dfrac{150}{2^3}\right\rfloor+ \left\lfloor\dfrac{150}{2^4}\right\rfloor+ \left\lfloor\dfrac{150}{2^5}\right\rfloor+ \left\lfloor\dfrac{150}{2^6}\right\rfloor+ \left\lfloor\dfrac{150}{2^7}\right\rfloor$, since $\left\lfloor\dfrac{150}{2^r}\right\rfloor = 0$ for any integer r >= 8.

Hence, $\max m = 146$. i.e., $2^{146}\Big|150!$, but $2^{147}\nmid 150!$.

Similarly, we have $\max k = 72$. i.e., $3^{72}\Big|150!$, but $3^{73}\nmid 150!$.

Now, we have 146 "two"s and 72 "three"s for us to pick from the prime factorization of 150!. To make 12, we take 2 "two"s and 1 "three". In total, we can make $\min\left(\dfrac{146}2, 72\right) = 72$ "twelve"s from what we have.

Therefore, the largest power of 12 such that 12^n divides 150! is $n = 72$.

Expand and move every term to one side.

$\begin{array}{rcl} (2x + 5)(x - 3) &=& 14 + 7x -8\\ 2x^2 + 5x - 6x - 15 &=& 14 + 7x - 8\\ 2x^2 -8x-7&=&0 \end{array}$

Let a and b be the possible values of x. Then by Vieta's formulae, $a + b = 4$.

Therefore, the sum of possible values of x is 4.

Move every term to one side:

$\begin{array}{rcl} x^2 -13x + 4&=& 5x + 7\\ x^2 - 18x - 3 &=& 0 \end{array}$

Let a and b be the roots. Then by Vieta's formulae,

$\begin{cases}a + b =18\\ab=-3\end{cases}$

Therefore, 

$\begin{array}{cl} &\text{sum of square of roots}\\ =& a^2 + b^2\\ =& (a + b)^2 - 2ab\\ =& 18^2 - 2(-3)\\ =& 330 \end{array}$

This is quite weird, because the area between $y=x - \dfrac{k^3}{x^2}$ and the x-axis is infinite, not 6.5 square units. I think there is something missing in the question, making it "unsolvable".

$\begin{array}{cl} &\sqrt{\dfrac{8!}{70}}\\ =& \sqrt{\dfrac{2\times 3\times 4\times 5\times 6\times 7\times 8}{2\times 5\times 7}}\\ =& \sqrt{3\times 4\times 6\times 8}\\ =& \sqrt{3\times 4 \times (2\times 3)\times (2\times 4)}\\ =& \sqrt{(2\times 2) \times ((3\times 4)\times (3\times 4))}\\ =& \sqrt{2^2 \times 12^2}\\ =& \sqrt{24^2}\\ =& 24 \end{array}$