Our first two terms each involve an exponent in parentheses raised to an exponent outside the parentheses, so in this case, for each term, we multiply the exponent inside the parentheses with the exponent outside the parentheses. For the third term, we have the same two numbers raised to a certain power, so we add the exponents. This gives us:

\(\displaystyle (2^2)^2+(3^2)^2+32(\frac{1}{2})^2(\frac{1}{2})^3\)

\(\displaystyle =2^4+3^4+32(\frac{1}{2})^5\)

We can then solve the expression:

\(\displaystyle =16+81+32(\frac{1}{32})\)

\(\displaystyle =16+81+1\)

\(\displaystyle =98\)

Recall the rule dealing with raising exponents to a higher power when an exponent in parenthesis is raised to an exponent outside of the parentheses; multiply the exponents together to get the new exponent. Don't be confused by the fractional exponents. Simply multiply across the numerator and across the denominator.

\(\displaystyle \small \small (z^{\frac{4}{3}})^{\frac{6}{8}}=z^\frac{4*6}{3*8}=z^{\frac{24}{24}}=z^1=z\)

In this case, the expression simplifies down to just \(\displaystyle z\).

For any positive integer whose last digit is \(\displaystyle a\), the last digit if this integer raised to any power \(\displaystyle p\) is the same as the last digit of \(\displaystyle a^{P}\). For our problem, the last digit is then given by \(\displaystyle 7^{43}\). Now, this is pretty complicated to calculate, so we can try to find a pattern in the last digits of the powers of 7.

\(\displaystyle 7^{1}\) is \(\displaystyle 7\)

\(\displaystyle 7^{2}\) is \(\displaystyle 9\)

\(\displaystyle 7^{3}\) is \(\displaystyle 3\)

\(\displaystyle 7^{4}\) is \(\displaystyle 1\)

\(\displaystyle 7^{5}\) is \(\displaystyle 7\) again.

\(\displaystyle 7^6\) is 9 again.

So, the pattern repeats every 4 numbers. Therefore, if we divide the power of 43 in \(\displaystyle 7^{43}\) by 4, we get a remainder of 3. Therefore, the final answer is the last digit of \(\displaystyle 7^{3}\), or \(\displaystyle 3\).

For any positive integer whose last digit is \(\displaystyle a\), the last digit if this integer raised to any power \(\displaystyle p\) is the same as the last digit of \(\displaystyle a^{P}\). So, the last digit of \(\displaystyle 1239^{56}\) is given by its last digit raised to the power of 56, or \(\displaystyle 9^{56}\).

Let's then try to find a pattern in the last digits of the powers of 9.

\(\displaystyle 9^{1}\) is \(\displaystyle 9\)

\(\displaystyle 9^{2}\) is \(\displaystyle 1\)

\(\displaystyle 9^{3}\) is \(\displaystyle 9\)

We can see that the odd powers of 9 have 9 as their last digits and the even powers of 9 have 1 for their last digits.

Since 56 is an even number, the last digit of \(\displaystyle 1239^{56}\) is \(\displaystyle 1\).

For any positive integer whose last digit is \(\displaystyle a\), the last digit if this integer raised to any power \(\displaystyle p\) is the same as the last digit of \(\displaystyle a^{P}\). So, the last digit of \(\displaystyle 763^{47}\) is the same as the last digit of \(\displaystyle 3^{47}\). Let's try to find a pattern in the last digits of the powers of 3:

For \(\displaystyle 3^{1}\), the last digit is 3

\(\displaystyle 3^{2}\) is 9

\(\displaystyle 3^{3}\) is 7

\(\displaystyle 3^{4}\) is 1

\(\displaystyle 3^{5}\) is 3

So the pattern repeats every 4 consecutive powers. We therefore simply have to divide 47 by 4 to get a remainder of 3. The last digit is given by \(\displaystyle 3^{3}\), or 7. To get the final answer, we simply have to multiply 7 by 8; we get 56, whose last digit is \(\displaystyle 6\), which is our final answer.

Some trinomials of the form \(\displaystyle x^{2}+Bx+ C\) are factorable as

\(\displaystyle (x+m)(x+n)\)

where \(\displaystyle mn = C\), \(\displaystyle m+n =B\).

Therefore, if the number in the circle makes the polynomial factorable, it can be factored as the product of two whole numbers whose sum is 17.

These numbers are:

\(\displaystyle 1 \times 16 = 16\)

\(\displaystyle 2 \times 15 = 30\)

\(\displaystyle 3 \times 14 = 42\)

\(\displaystyle 4 \times 13 = 52\)

\(\displaystyle 5 \times 12 = 60\)

\(\displaystyle 6 \times 11 = 66\)

\(\displaystyle 7 \times 10 = 70\)

\(\displaystyle 8 \times 9 = 72\)

Of the five choices, only 56 is not among these numbers, so it makes the polynomial prime. It is the correct choice.

Some trinomials of the form \(\displaystyle x^{2}-Bx- C\) are factorable as

\(\displaystyle (x+m)(x-n)\)

where \(\displaystyle mn = C\), \(\displaystyle n - m = B\).

Therefore, we are looking for an integer which can be factored as the product of two integers whose difference is 9.

We can start with 1 and 10 and work our way up:

\(\displaystyle 10 \times 1 = 10\)

\(\displaystyle 11 \times 2 = 22\)

\(\displaystyle 12 \times 3 = 36\)

\(\displaystyle ...\)

The products will increase, so we can stop, having found our two numbers, 2 and 11. Their product, 22, is the correct choice.

If \(\displaystyle x^{2}+ B x- 60\) is not prime, then, as a quadratic trinomial of the form \(\displaystyle x^{2}+ B x- C\) it is factorable as 

\(\displaystyle (x+m)(x-n)\)

where \(\displaystyle mn = 60\) and \(\displaystyle m - n = B\).

Therefore, we are looking for a whole number that is not the difference of two factors of 60. The integers that are such a difference are

\(\displaystyle 60 -1= 59\)

\(\displaystyle 30-2=28\)

\(\displaystyle 20-3=17\)

\(\displaystyle 15-4=11\)

\(\displaystyle 12-5 = 7\)

\(\displaystyle 10 -6 = 4\)

Of the choices, only 23 is not on the list, so it is the correct choice.

The factoring pattern to look for in the second polynomial is the sum of cubes, so the number in the circle must be a perfect cube of a whole number \(\displaystyle B\). We can write this polynomial as

\(\displaystyle x^{3}+ B^{3}\)

which can be factored as 

\(\displaystyle (x+B) (x^{2}-Bx+B^{2})\)

By the condition of the problem, \(\displaystyle B = 8\), so the number that replaces the square is \(\displaystyle 8^{2} = 64\), and the number that replaces the circle is \(\displaystyle 8^{3} = 512\).

There are different ways to approach this problem. We just need to remember three things:

1. \(\displaystyle \frac{x^a}{x^b}=x^{a-b}\)
2. \(\displaystyle x^{-a}=\frac{1}{x^a}\)
3. \(\displaystyle x^ax^b=x^{a+b}\)

Keeping those in mind, we can simplify the numerator and coefficients:

\(\displaystyle \frac{3x^{-4}y^2zx^2}{21yz^{-3}}=\frac{1x^{-4+2}y^2z}{7yz^{-3}}=\frac{1x^{-2}y^2z}{7yz^{-3}}\)

I'm going to move the negative exponents (number 2 in the list above) in order to make them positive: 

\(\displaystyle \frac{1\mathbf{x^{-2}}y^2z}{7y\mathbf{z^{-3}}}=\frac{z^3y^2z}{7x^2y}\)

We can now simplify the numerator again with the \(\displaystyle z\) exponents (number 3) and the \(\displaystyle y\) exponents (number 1): 

\(\displaystyle \frac{z^3y^2z}{7x^2y}=\frac{yz^4}{7x^2}\)