We can rewrite as follows:

\(\displaystyle y^{4} - x^{2} + 8x - 16\)

\(\displaystyle = y^{4} - (x^{2} - 8x + 16)\)

\(\displaystyle y^{4} = \left (y ^{2} \right )^{2}\), and 

\(\displaystyle x^{2} - 8x + 16\) is a perfect square polynomial, as seen here:

\(\displaystyle x^{2} - 8x + 16 = x^{2} - 2 \cdot 4 \cdot x + 4^{2} = (x -4)^{2}\)

so the original polynomial is equal to 

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

This is the difference of squares, so it can be factored as

\(\displaystyle \left [y ^{2} + (x -4) \right ]\left [y ^{2} - (x -4) \right ]\)

\(\displaystyle = (y ^{2}+ x -4) (y ^{2}- x +4)\)

The easiest way to solve this problem is to take a negative integer to use for m. 

For example, \(\displaystyle -2\) can be used. 

Plugging in \(\displaystyle -2\) into the expression, \(\displaystyle 5m^{2}-2m\), we get:

\(\displaystyle 5\cdot (-2^{2})-2\cdot (-2)\)

This simplifies to 

\(\displaystyle 5\cdot 4+4\)

\(\displaystyle 20+4\)

\(\displaystyle 24\)

Given that 24 is a positive number, \(\displaystyle 5m^{2}-2m\) is the correct answer. 

By the order of operations, all operations within grouping symbols must be performed first, with the innermost symbols taking precedence. Therefore, the three operations within the brackets - the subtraction, the division, and the cubing - must be performed before the remaining two.

Once these three operations are completed, there remain two more, a division and an addition. Division has precendence in the order of operations, so the last operation performed is the addition.

Since 1 mile is equivalent to 1.609 kilometers, the number of kilometers equivalent to 100 miles can be found by multiplying 100 by 1.609. Conversely, the number of miles equivalent to 100 kilometers can be found by dividing 100 by 1.609. 

You do not have to do the actual math to answer the question. Since the conversion factor is greater than one, multiplying any positive number by this factor yields a result greater than dividing that same number by it. Therefore, 

\(\displaystyle 100 \times 1.609 > 100 \div 1.609\),

and the number of kilometers equivalent to 100 miles, (b), is the greater quantity.

One foot comprises twelve inches, so multiply the number of feet by conversion factor 12:

\(\displaystyle 16x \cdot 12 = 192x\) inches.

One pound comprises sixteen ounces, so multiply the number of ounces by conversion factor 16:

\(\displaystyle 12x \cdot 16 = 192x\) ounces.

The two quantities are both equal to \(\displaystyle 192x\).

One yard comprises thirty-six inches, so multiply the number of yards by conversion factor 36:

\(\displaystyle \frac{7}{36}x \cdot 36 = 7x\)

One week comprises seven days, so multiply the number of weeks by conversion factor 7:

\(\displaystyle \frac{36} {7} x \cdot 7 = 36x\)

(b) is the greater quantity, since, if \(\displaystyle x\) is positive, \(\displaystyle 36x > 7x\).

\(\displaystyle (2t+u) (2t-u)\) is the product of the sum and the difference of the same two binomials, so this can be rewritten, and evaluated, using the difference of squares pattern:

\(\displaystyle (2t+u) (2t-u)\)

\(\displaystyle = (2 t)^{2} - u^{2}\)

\(\displaystyle = 2^{2} \cdot t ^{2} - u^{2}\)

\(\displaystyle = 4 t ^{2} - u^{2}\)

\(\displaystyle = 4 (40 ) - 17\)

\(\displaystyle = 160 - 17\)

\(\displaystyle = 143\)

Another way to express ratios is through division. 10 divided by 14 is approximate 0.71.

A quotient is the result of division. If \(\displaystyle q\) is the quotient of \(\displaystyle p\) and \(\displaystyle v\), that means that \(\displaystyle p\div v=q\) could be true.

Since the quotient of negative numbers is positive, both results will be positive.

We can rewrite both of these as products of positive numbers, as follows:

\(\displaystyle A \div \left ( - \frac{1}{3}\right ) = \left | A \right | \div \frac{1}{3} = \left | A \right | \cdot 3\)

\(\displaystyle A \div \left ( -0.3\right ) =A \div \left ( - \frac{3}{10}\right ) = \left | A \right | \div \frac{3}{10} = \left | A \right | \cdot \frac{10}{3} = \left | A \right | \cdot 3 \frac{1}{3}\)

\(\displaystyle 3 \frac{1}{3} > 3\), so

\(\displaystyle \left | A \right | \cdot 3 \frac{1}{3} >\left | A \right | \cdot 3\), and 

\(\displaystyle A \div \left ( -0.3\right ) > A \div \left ( - \frac{1}{3}\right )\)

making (B) greater.