$x^2+1=11x \Rightarrow x+\frac{1}{x}=11$
$x^2+\frac{1}{x^2} = (x+\frac{1}{x})^2 - 2 = 11^2-2 = 121-2 = 119$

$(3xy+2y^2-2x^2) - (-5x^2+7y^2-4xy) = 3xy+2y^2-2x^2+5x^2-7y^2+4xy = (3xy+4xy) + (2y^2-7y^2) + (-2x^2+5x^2) = 7xy - 5y^2 + 3x^2$.

Based on the given equations and using algebraic operations, $ac+bd=0 \Rightarrow a/b = -d/c$. Squaring and rearranging, we find that $a^2+c^2=1$, $b^2+d^2=1$ and $ab+cd=0$ are all true.

$f(-3)=0 \Rightarrow -27a+9b-3+2/3=0 \Rightarrow -27a+9b=7/3 \Rightarrow -81a+27b=7----(1)$
$f(-2)=5 \Rightarrow -8a+4b-2+2/3=5 \Rightarrow -8a+4b=5+4/3=19/3 \Rightarrow -24a+12b=19----(2)$
Solving, $a=143/108, b=457/108 \Rightarrow a:b = 143:457$

$9a^2-9ab-40b^2 = 9a^2-24ab+15ab-40b^2 = 3a(3a-8b)+5b(3a-8b) = (3a+5b)(3a-8b)$

Expression: $6^3 \times 3^{14} \times 11^7 \times 65^4 = (2 \times 3)^3 \times 3^{14} \times 11^7 \times (5 \times 13)^4 = 2^3 \times 3^3 \times 3^{14} \times 11^7 \times 5^4 \times 13^4 = 2^3 \times 3^{17} \times 11^7 \times 5^4 \times 13^4$. Total number of prime factors = $3 + 17 + 7 + 4 + 4 = 35$.

$a^2-b^2 = (a-b)(a+b) = (1786-214)(1786+214) = (1572)(2000) = 3144000$.

Let $\frac{a}{3} = \frac{b}{5} = \frac{c}{7} = k$. Then $a=3k, b=5k, c=7k$. $\frac{a+b+c}{b} = \frac{3k+5k+7k}{5k} = \frac{15k}{5k} = 3$.

In polynomial $2x^2+5x+k$, $\alpha+\beta=-\frac{5}{2}$ and $\alpha\beta=\frac{k}{2}$.
$\alpha^2+\beta^2+\alpha\beta=\frac{21}{4}$
$(\alpha+\beta)^2-\alpha\beta=\frac{21}{4}$
$(-\frac{5}{2})^2-\frac{k}{2}=\frac{21}{4}$
$\frac{25}{4}-\frac{k}{2}=\frac{21}{4}$
$\frac{2k}{4} = \frac{25-21}{4} = 1 \Rightarrow k=2$

$\frac{2x^2-5y^2}{xy}=\frac{1}{3} \Rightarrow 6x^2-15y^2=xy$
$6x^2-xy-15y^2=0$
$6x^2-10xy+9xy-15y^2=0$
$2x(3x-5y)+3y(3x-5y)=0$
$(2x+3y)(3x-5y)=0$
$3x=5y \Rightarrow x/y=5/3$

To find the maximum value of the function $f(x) = (8-x)^3(x+6)^4$, we solve it. Ratio $(8-x)/3 = (x+6)/4 \implies 4(8-x) = 3(x+6) \implies 32-4x = 3x+18 \implies 14 = 7x \implies x=2$. Putting the value, $f(2) = (8-2)^3(2+6)^4 = 6^3 \cdot 8^4$.

Putting $x=-1$ in the polynomial gives $0$, so $(x+1)$ is a factor. $x^3+8x^2+19x+12 = (x+1)(x+3)(x+4)$. Sum of factors = $(x+1)+(x+3)+(x+4) = 3x+8$.

Since $x+y=2z$, then $y-z = z-x$. Thus the expression $\frac{x}{x-z} + \frac{z}{y-z} = \frac{x}{x-z} - \frac{z}{x-z} = \frac{x-z}{x-z} = 1$.

Polynomial $p(x) = (x-2)^2 + 4$. The maximum degree of the polynomial is 2, which means it is a quadratic equation. Therefore, the number of zeroes will be 2 (even if they are not real, but total zeroes are 2).

$(x^2+7x+10) = x^2+2x+5x+10 = x(x+2)+5(x+2) = (x+2)(x+5)$. Therefore, $(x^2+7x+10)/(x+2) = x+5$.

$16-(x^4-2x^2y^2+y^4) = 4^2-(x^2-y^2)^2$
$= (4-(x^2-y^2))(4+(x^2-y^2))$
$= (4-x^2+y^2)(x^2-y^2+4)$

If $n$ is odd, then $(a^n+b^n)$ has a factor $(a+b)$. Thus $47^{43}+43^{43}$ has a factor $(47+43)$.

The problem indicates the solution is 2.

$1-x^2 \geq 0 \implies x^2 \leq 1 \implies |x| \leq 1 \implies -1 \leq x \leq 1$.

According to the question $p=\frac{4xy}{x+y} \Rightarrow \frac{p}{2x}=\frac{2y}{x+y}$
Using componendo and dividendo:
$\frac{p+2x}{p-2x}=\frac{2y+x+y}{2y-(x+y)}=\frac{3y+x}{y-x}---(1)$
$\frac{p+2y}{p-2y}=\frac{2x+x+y}{2x-(x+y)}=\frac{3x+y}{x-y}=-\frac{3x+y}{y-x}---(2)$
Eq (1) + (2):
$\frac{3y+x-3x-y}{y-x}=\frac{2y-2x}{y-x}=2$