$\displaystyle \small (x-4)(x+2)(2x-5)$

When multiplying, the order in which you multiply does not matter. Let's start with the first two monomials.

 $\displaystyle (x-4)(x+2)$

Use FOIL to expand.

$\displaystyle x^2+2x-4x-8=x^2-2x-8$

Now we need to multiply the third monomial.

$\displaystyle \small (x-4)(x+2)(2x-5)=(x^2-2x-8)(2x-5)$

Similar to FOIL, we need to multiply each combination of terms.

$\displaystyle 2x(x^2-2x-8)+(-5)(x^2-2x-8)$

$\displaystyle 2x^3-4x^2-16x-5x^2+10x+40$

Combine like terms.

$\displaystyle 2x^3-9x^2-6x+40$

You can look at this as the sum of two expressions multiplied by the difference of the same two expressions. Use the pattern

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

where $\displaystyle \small A = x^{2} - 2x$ and $\displaystyle \small B = 7$.

$\displaystyle \small \small \small (x^{2} - 2x + 7) (x^{2} - 2x - 7) = (x^{2} - 2x )^{2} - 7^{2} = (x^{2} - 2x )^{2} - 49$ 

To find $\displaystyle \small (x^{2} - 2x )^{2}$, you use the formula for perfect squares:

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

where $\displaystyle \small \small A = x^{2}$ and $\displaystyle \small B = 2x$.

$\displaystyle \small \small \small (x^{2}-2x)^{2} = ((x^{2})^{2}-2x^{2}(2x)+(2x)^{2}) = x^{4}-4x^{3}+4x^2$

Substituting above, the final answer is $\displaystyle \small x^{4}-4x^{3}+4x^2-49$ .

A polynomial of the form $\displaystyle ax^{2} +bx + c$ whose terms do not have a common factor, such as this, can be factored by rewriting it as $\displaystyle ax^{2} +fx + gx + c$ such that $\displaystyle f + g =b$ and $\displaystyle fg = ac$; the grouping method can be used on this new polynomial.

Therefore, for  $\displaystyle 3x^{2} +Nx + 32$ to be factorable, $\displaystyle N$ must be the sum of the two integers of a factor pair of $\displaystyle 3 \times 32 = 96$. We are looking for a value of $\displaystyle N$ that is not a sum of two such factors.

The factor pairs of 96, along with their sums, are:

1 and 96 - sum 97

2 and 48 - sum 50

3 and 32 - sum 35

4 and 24 - sum 28

6 and 16 - sum 22

8 and 12 - sum 20

Of the given choices, only 30 does not appear among these sums; it is the correct choice.

$\displaystyle y^{3} - 125 = y^{3} - 5 ^{3}$

making this polynomial the difference of two cubes.

As such,  $\displaystyle y^{3} - 125$ can be factored using the pattern

$\displaystyle A ^{3} - B ^{3} = (A - B) (A^{2}+ AB + B^{2})$

so

$\displaystyle y^{3} - 125 = y^{3} - 5 ^{3} = (y - 5) (y^{2}+ y \cdot 5 + 5^{2})= (y - 5) (y^{2}+ 5y + 2 5 )$

(A) and (C) are both factors, but not (B) or (D), so the correct response is two.

$\displaystyle \left (\frac{1}{3}x^{2}+\frac{1}{6}xy-\frac{5}{4}y^{2} \right )-\left (\frac{1}{9}x^{2}-\frac{4}{3}xy+y^{2} \right )$

Since we are only adding and subtracting (there is no multiplication or division), we can remove the parentheses.

$\displaystyle \frac{1}{3}x^{2}+\frac{1}{6}xy-\frac{5}{4}y^{2}-\left \frac{1}{9}x^{2}+\frac{4}{3}xy-y^{2} \right$

Regroup the expression so that like variables are together. Remember to carry positive and negative signs.

$\displaystyle (\frac{1}{3}x^{2}-\left \frac{1}{9}x^{2})+(\frac{1}{6}xy+\frac{4}{3}xy)-(\frac{5}{4}y^{2}-y^{2} \right)$

For all fractional terms, find the least common multiple in order to add and subtract the fractions.

$\displaystyle (\frac{3}{9}x^{2}-\left \frac{1}{9}x^{2})+(\frac{1}{6}xy+\frac{8}{6}xy)-(\frac{5}{4}y^{2}-\frac{4}{4}y^{2} \right)$

Combine like terms and simplify.

$\displaystyle \frac{2}{9}x^{2}+\frac{9}{6}xy-\frac{9}{4}y^{2}$

$\displaystyle \frac{2}{9}x^{2}+\frac{3}{2}xy-\frac{9}{4}y^{2}$

First, set up the division as the following:

$\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \end{array}$

Look at the leading term $\displaystyle x^2$ in the divisor and $\displaystyle 3x^3$ in the dividend. Divide $\displaystyle 3x^3$ by $\displaystyle x^2$ gives $\displaystyle 3x$; therefore, put $\displaystyle 3x$ on the top:

$\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \end{array}$

Then take that $\displaystyle 3x$ and multiply it by the divisor, $\displaystyle x^2-1$, to get $\displaystyle 3x^3-3x$.  Place that $\displaystyle 3x^3-3x$ under the division sign:

$\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \end{array}$

Subtract the dividend by that same $\displaystyle 3x^3-3x$ and place the result at the bottom. The new result is $\displaystyle 5x^2+2x+8$, which is the new dividend.

$\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \end{array}$

Now, $\displaystyle 5x^2$ is the new leading term of the dividend.  Dividing $\displaystyle 5x^2$ by $\displaystyle x^2$ gives 5.  Therefore, put 5 on top:

$\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \end{array}$

Multiply that 5 by the divisor and place the result, $\displaystyle 5x^2-5$, at the bottom:

$\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \\ & && & 5x^2 & -5 \end{array}$

Perform the usual subtraction:

$\displaystyle \begin{array}{*2r @{\hskip\arraycolsep}c@{\hskip\arraycolsep} *4r} & && 3x & 5 & & \\ \cline{3-7} x^2 & -1 &\big)& 3x^3 & 5x^2 & -x & 8 \\ & && 3x^3 & -3x \\ \cline{4-7} & && & 5x^2 & 2x & 8 \\ & && & 5x^2 & -5 \\ \cline{5-7} & && & & 2x & 13 \\ \end{array}$

Therefore the answer is $\displaystyle 3x+5$ with a remainder of $\displaystyle 2x+13$, or $\displaystyle 3x+5+\frac{2x+13}{x^2-1}$.

$\displaystyle n^{4} - 32n^{2} +256$ can be seen to fit the pattern 

$\displaystyle A ^{2} - 2AB + B^{2}$:

$\displaystyle n^{4} - 32n^{2} +256 = \left (n^{2} \right ) ^{2}- 2 \cdot n^{2} \cdot 16 + 16 ^{2}$

where $\displaystyle A = n^{2} , B = 16$

$\displaystyle A ^{2} - 2AB + B^{2}$ can be factored as $\displaystyle (A-B ) ^{2}$, so 

$\displaystyle n^{4} - 32n^{2} +256 = \left (n^{2} \right ) ^{2}- 2 \cdot n^{2} \cdot 16 + 16 ^{2} =\left ( n ^{2} - 16 \right )^{2}$

$\displaystyle n^{2} - 16 = n^{2} - 4^{2}$, making this the difference of squares, so it can be factored as follows:

$\displaystyle n^{2} - 16 = n^{2} - 4^{2} = (n + 4) (n - 4)$

Therefore, 

$\displaystyle n^{4} - 32n^{2} +256 = \left ( n ^{2} - 16 \right )^{2} = (n+4) ^{2} (n-4) ^{2}$

The polynomial has only two prime factors, each squared, neither of which appear among the choices.

Divide termwise:

$\displaystyle \left ( 40x^{3} - 24x^{2}+ 20x- 64\right ) \div 8 x^{2}$

$\displaystyle = \frac{40x^{3} - 24x^{2}+ 20x- 64}{ 8 x^{2}}$

$\displaystyle = \frac{40x^{3} }{ 8 x^{2}}- \frac{ 24x^{2}}{ 8 x^{2}}+ \frac{ 20x}{ 8 x^{2}}- \frac{ 64}{ 8 x^{2}}$

$\displaystyle = \frac{40 }{ 8}x^{3-2}- \frac{ 24}{ 8 }+ \frac{ 20x}{ 8 } \cdot \frac{1}{x^{2-1}}- \frac{ 64}{8 }\cdot \frac{1}{x^{2} }$

$\displaystyle =5x-3+ \frac{ 5}{ 2 } \cdot \frac{1}{x}- 8 \cdot \frac{1}{x^{2} }$

$\displaystyle =5x-3+ \frac{ 5}{ 2x } - \frac{8}{x^{2} }$

$\displaystyle 27N^{3}-1,331$ can be rewritten as $\displaystyle (3N)^{3} - 11^{3}$ and is therefore the difference of two cubes. As such, it can be factored using the pattern

$\displaystyle A^{3}-B^{3} =(A-B) (A^{2}+AB +B^{2})$

where $\displaystyle A = 3N, B = 11$.

$\displaystyle 27N^{3}-1,331$

$\displaystyle =(3N)^{3} - 11^{3}$

$\displaystyle = (3N-11)[(3N)^{2}+3N\cdot 11 +11^{2}]$

$\displaystyle = (3N-11)(9N^{2}+33N +121)$

Since both terms are perfect cubes $\displaystyle \left (512= 8^{3} \right )$, the factoring pattern we are looking to take advantage of is the sum of cubes pattern. This pattern is

$\displaystyle A^{3}+ B^{3} = (A + B) (A^{2}-AB +B^{2})$

We substitute $\displaystyle t$ for $\displaystyle A$ and 8 for $\displaystyle B$:

$\displaystyle t^{3} + 512= t^{3} + 8^{3}$

$\displaystyle = (t+ 8) (t^{2}-t \cdot 8+8^{2})$

$\displaystyle = (t+ 8) (t^{2}-8t+64)$