The polynomial is the difference of squares and can be factored using the pattern 

$\displaystyle A^{2} - B ^{2} = (A+B)(A-B)$

where 

$\displaystyle A = 6x, B = 7y$

as seen here:

$\displaystyle 36x^{2} - 49y^{2}$

$\displaystyle = 6^{2}x^{2} - 7^{2}y^{2}$

$\displaystyle = \left (6 x \right )^{2} - \left (7 y \right )^{2}$

$\displaystyle = \left (6x + 7y \right )\left (6x - 7y \right )$

The greatest common factor of the two terms is the monomial term $\displaystyle 4x ^{2}y$, so factor it out:

$\displaystyle 8x ^{3}y - 20x ^{2}y ^{2}$

$\displaystyle = 4x ^{2}y \cdot 2x - 4x ^{2}y \cdot 5y$

$\displaystyle = 4x ^{2}y \left (2x - 5y \right )$

Of the four choices, $\displaystyle 2x - 5y$ is correct.

The greatest common factor of the two terms is the monomial term $\displaystyle 4x ^{2}y$, so factor it out:

$\displaystyle 8x ^{3}y - 20x ^{2}y ^{2}$

$\displaystyle = 4x ^{2}y \cdot 2x - 4x ^{2}y \cdot 5y$

$\displaystyle = 4x ^{2}y \left (2x - 5y \right )$

Of the four choices, $\displaystyle 2x - 5y$ is correct.

Raise a fraction to a negative power by raising its reciprocal to the power of the absolute value of the exponent. Then apply the power of a quotient rule:

$\displaystyle \left (\frac{5}{t} \right )^{-3} = \left (\frac{t}{5} \right )^{ 3} = \frac{t^{ 3}}{5^{ 3}} = \frac{t^{ 3}}{125}$

To raise a number to a negative exponent, raise it to the absolute value of that exponent, then take its reciprocal. We do this, then apply the various properties of exponents:

$\displaystyle \left (4 t^{2} \right )^{-3} = \frac{1}{\left (4 t^{2} \right )^{3} } = \frac{1}{4^{3} \left ( t^{2} \right )^{3} } =\frac{1}{64 t^{2 \cdot 3} } =\frac{1}{64 t^{6} }$

For a quadratic trinomial with a quadratic coefficient other than 1, use the factoring by grouping method.

First, find two integers whose product is $\displaystyle 6 \times (-15) = -90$ (the product of the quadratic and constant coefficients) and whose sum is 1 (the implied coefficient of $\displaystyle t$ ). By trial and error, we find that these are $\displaystyle -9, 10$. 

Split the linear term accordingly, then factor by grouping, as follows.

$\displaystyle 6t ^{2} + t - 15$

$\displaystyle = 6t ^{2} - 9t +10 t - 15$

$\displaystyle =\left ( 6t ^{2} - 9t \right )+\left (10 t - 15 \right )$

$\displaystyle = \left ( 3t \cdot 2t - 3t \cdot 3 \right )+\left (5 \cdot 2t - 5 \cdot 3 \right )$

$\displaystyle = 3t \left ( 2t - 3 \right )+ 5 \left ( 2t - 3 \right )$

$\displaystyle =\left ( 3t + 5 \right )\left ( 2t - 3 \right )$

The greatest common factor of the terms is $\displaystyle 4y$, so factor it out:

$\displaystyle 4y^{3} - 12y ^{2}- 72y$

$\displaystyle =4y \cdot y^{2} - 4y \cdot 3y - 4y \cdot 18$

$\displaystyle =4y \left ( y^{2} - 3y - 18 \right )$

The trinomial might be able to be factored as 

$\displaystyle (y+A)(y + B)$,

where $\displaystyle AB = -18$ and $\displaystyle A + B = -3$.

By trial and error, we find that 

$\displaystyle A = -6 , B = 3$,

so the factorization becomes

$\displaystyle 4y (y-6)(y+3)$.

A number decreased by 40% is equivalent to 100% of the number minus 40% of the number. This is taking 60% of the number, or, equivalently, multiplying it by 0.6. 

Therefore, $\displaystyle 8t + 72$ decreased by 40% is 0.6 times this, or

$\displaystyle 0.6 \left (8t + 72 \right ) =4.8t + 43.2$.

A number increased by 20% is equivalent to 100% of the number plus 20% of the number. This is taking 120% of the number, or, equivalently, multiplying it by 1.2.

Therefore, $\displaystyle 8x + 9y$ increased by 20% is 1.2 times this, or

$\displaystyle 1.2 \left (8x + 9y \right ) = 1.2 \cdot 8x + 1.2 \cdot 9y = 9.6x + 10.8 y$.

This can be most easily solved by first substituting $\displaystyle t$ for $\displaystyle n^{2}$, and, subsequently, $\displaystyle t^{2}$ for $\displaystyle n^{4}$:

$\displaystyle n^{4} - 16n^{2} - 225$

$\displaystyle = t^{2} - 16t - 225$

This becomes quadratic in the new variable, and can be factored as

$\displaystyle \left (t +\; \; \right )\left (t +\; \; \right )$,

filling out the blanks with two numbers whose sum is $\displaystyle -16$ and whose product is $\displaystyle -225$. Through some trial and error, the numbers can be seen to be $\displaystyle 9, -25$.

Therefore, after factoring and substituting back,

$\displaystyle n^{4} - 16n^{2} - 225$

$\displaystyle = t^{2} - 16t - 225$

$\displaystyle = (t+9)(t-25)$

$\displaystyle = (n^{2}+9)(n^{2}-25)$

The first factor, the sum of squares, is prime. The second factors as the difference of squares, so the final factorization is

$\displaystyle (n^{2}+9)(n+5)(n-5)$.

Of the choices given, $\displaystyle n^{2}+9$ is correct.