Simplify the following expressions fully. Leave your answers with a positive exponent where necessary:

Distribute the $-2$ to each term inside the parentheses.

The simplified expression is $\boxed{-2x+14}$.

Expand the binomial squared using the formula $(a-b)^2 = a^2 - 2ab + b^2$.
Here, $a=3x$ and $b=5$.

The simplified expression is $\boxed{9x^2-30x+25}$.

Recognize this as the difference of cubes formula: $(a-b)(a^2+ab+b^2) = a^3 - b^3$.
Here, $a=x$ and $b=3$.

The simplified expression is $\boxed{x^3-27}$.

Recognize this as the difference of squares formula: $(a+b)(a-b) = a^2 - b^2$.
Here, $a=\frac{p}{5}$ and $b=1$.

The simplified expression is $\boxed{\frac{p^2}{25}-1}$.

Find the greatest common factor (GCF) of $3p$ and 15, which is 3.

The factored expression is $\boxed{3(p-5)}$.

Factor the quadratic trinomial $2x^2-3x-5$.
We look for two numbers that multiply to $2 \times -5 = -10$ and add to $-3$. These numbers are $-5$ and 2.
Rewrite the middle term using these numbers:

Group the terms and factor:

The factored expression is $\boxed{(x+1)(2x-5)}$.

Group the first three terms, which form a perfect square trinomial, and then apply the difference of squares formula.

Now apply the difference of squares formula, $X^2 - Y^2 = (X-Y)(X+Y)$, where $X=(a-b)$ and $Y=8c$.

The factored expression is $\boxed{(a-b-8c)(a-b+8c)}$.

Given $m = \sqrt{b^2 - 4ac}$ and $a=2, b=-3, c=-1$.
Substitute the values into the formula:

The value of $m$ in simplest surd form is $\boxed{\sqrt{17}}$.

Since $m = \sqrt{17}$ and 17 is not a perfect square, $\sqrt{17}$ cannot be expressed as a simple fraction of two integers.
Therefore, $m$ is irrational.
The value of $m$ is $\boxed{\text{irrational}}$.

We know $m = \sqrt{17}$.
We need to find two consecutive integers whose squares are just below and just above 17.
$4^2 = 16$
$5^2 = 25$
Since $16 < 17 < 25$, it follows that $\sqrt{16} < \sqrt{17} < \sqrt{25}$.
So, $4 < \sqrt{17} < 5$.
Thus, $m$ lies between the integers $\boxed{4 \text{ and } 5}$.

Add 11 to both sides:

Divide by 3:

The solution is $\boxed{x=\frac{11}{3}}$.

Rearrange the equation to the standard quadratic form $ax^2+bx+c=0$:

Factor the quadratic equation. We look for two numbers that multiply to $2 \times -6 = -12$ and add to $-1$. These numbers are $-4$ and 3.
Rewrite the middle term:

Group the terms and factor:

Set each factor to zero to find the solutions:

The solutions are $\boxed{x=-\frac{3}{2} \text{ or } x=2}$.

Express the right side with a base of 5:

So, $\frac{1}{125} = \frac{1}{5^3} = 5^{-3}$.
Now the equation is:

Since the bases are the same, the exponents must be equal:

The solution is $\boxed{x=-3}$.

First, distribute the 3 on the right side:

To eliminate the fraction, multiply the entire inequality by 2:

Subtract $x$ from both sides:

Subtract 42 from both sides:

Divide by 5:

The solution to the inequality is $\boxed{x > -8}$.

Draw a number line. Mark $-8$ on it. Since $x > -8$, use an open circle at $-8$ (because $x$ is strictly greater than $-8$, not equal to it) and draw an arrow extending to the right from $-8$.

<--------------------------------------------------------------------->
-10   -9   -8   -7   -6   -5   -4   -3   -2   -1    0    1    2    3
             o----------------------------------------------------->

The number line representation shows an open circle at $-8$ and a line extending to the right.