To factor $\displaystyle x^{2} + 12x + 24$ , we use the "reverse-FOIL".

$\displaystyle x^{2} + 12x + 24 = (x+A)(x+B )$, where $\displaystyle A$ and $\displaystyle B$ are two integers whose product is 24 and whose sum is 12. If we examine all of the factor pairs of 24, we note that the sum of each pair is as follows:

1 and 24: 25
2 and 12: 14
3 and 8: 11
4 and 6: 10

No factor pair of 24 has sum 12, so $\displaystyle x^{2} + 12x + 24$ cannot be factored. The correct response is that none of the four binomials is correct.

The sum of two squares is in general a prime polynomial unless a greatest common factor can be distributed out. $\displaystyle x^{2}+ 81$ is the sum of squares, and its terms do not have a GCF, so it is the prime polynomial.

Of the remaining choices:

$\displaystyle x^{2}-81$ is equal to $\displaystyle x^{2} - 9^{2}$; as the difference of squares, it is factorable.

$\displaystyle x^{2} - 18 x + 81$ is equal to $\displaystyle x^{2} - 2 \cdot x \cdot 9 + 9^{2}$; this fits the pattern of a perfect square quadratic trinomial, and is therefore factorable. $\displaystyle x^{2} +18 x + 81$ is factorable for a similar reason.

Let $\displaystyle A$ be the integer written in the box.

If $\displaystyle x^{2}+A x - 7$ is factorable, then its factorization is $\displaystyle (x+M)(x-N)$, where $\displaystyle MN = 7$ and $\displaystyle M - N = A$. In other words, $\displaystyle A$ must be the positive difference of two numbers of a factor pair of 7. 7 has only one factor pair, 1 and 7, so the only possible value of $\displaystyle A$ is $\displaystyle 7 - 1= 6$. The correct choice is one.

$\displaystyle x^{3}- 6x^{2}+ 12 x - 72$ can be factored by grouping, as follows:

$\displaystyle x^{3}- 6x^{2}+ 12 x - 72$

$\displaystyle =(x^{3}- 6x^{2})+( 12 x - 72)$

$\displaystyle =x^{2} (x - 6 )+12 (x - 6 )$

$\displaystyle =(x^{2} +12 )(x - 6 )$

$\displaystyle x^{2}+ 12$ cannot be factored further. Of the five choices, $\displaystyle x- 6$ is the only factor.

A quadratic trinomial of the form $\displaystyle a x^{2}+ bx + c$ can be factored by splitting the middle term into two terms. The two coefficients must have sum $\displaystyle b$ and product $\displaystyle ac$.

In each case, a factor pair of $\displaystyle 3 \times 8 = 24$ must be examined. These pairs, along with their sums, are:

1 and 24: 25
2 and 12: 14
3 and 8: 11
4 and 6: 10

Each polynomial with one of these four integers as its linear coefficient can be factored. The odd one out is $\displaystyle 3x^{2} +20x + 8$, since there is no factor pair of 24 whose sum is 20. This is the correct choice.

First, factor the numerator, which should be $\displaystyle (3x-5)(x-2)$.  Now the left side of your equation looks like 

$\displaystyle \frac{(3x-5)(x-2)}{x-2}$ 

Second, cancel the "like" terms - $\displaystyle (x-2)$ - which leaves us with $\displaystyle 3x-5$.  

Third, solve for $\displaystyle x$ by setting the left-over factor equal to 0, which leaves you with 

$\displaystyle 3x-5 =0$

$\displaystyle 3x=5$

$\displaystyle x=\frac{5}{3}$

Here you have an expression with three variables. To factor, you will need to pull out the greatest common factor that each term has in common.

Only the last two terms have $\displaystyle c$ so it will not be factored out. Each term has at least $\displaystyle a^2$ and $\displaystyle b$ so both of those can be factored out, outside of the parentheses. You'll fill in each term inside the parentheses with what the greatest common factor needs to be multiplied by to get the original term from the original polynomial:

$\displaystyle a^3b^2 + a^4b + a^2bc^3 - a^3b^3c^2 = a^2b(ab + a^2 + c^3 - ab^2c^2)$

To find the greatest common factor, we need to break each term into its prime factors:

$\displaystyle \small 2xy^3=2 \cdot x \cdot y \cdot y \cdot y$

$\displaystyle \small 14xy^2 = 2 \cdot 7 \cdot x \cdot y \cdot y$

$\displaystyle \small 9xy = 3 \cdot 3 \cdot x \cdot y$

Looking at which terms all three expressions have in common; thus, the GCF is $\displaystyle \small xy$. We then factor this out: $\displaystyle \small \small xy(2y^2 -14y +9)$. 

To find the greatest common factor, we must break each term into its prime factors:

$\displaystyle \small x^2y^3z^2 = x \cdot x \cdot y \cdot y \cdot y \cdot z \cdot z$

$\displaystyle \small x^4y^3z= x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot z$

The terms have $\displaystyle \small x^2$, $\displaystyle \small y^3$, and $\displaystyle \small z$ in common; thus, the GCF is $\displaystyle \small x^2y^3z$.

Pull this out of the expression to find the answer: $\displaystyle \small x^2y^3z(z+x^2)$. 

Consider the possible values for (x, y):

(1, 100)

(2, 50)

(4, 25)

(5, 20)

Note that (10, 10) is not possible since the two variables must be distinct. The sums of the above pairs, respectively, are:

1 + 100 = 101

2 + 50 = 52

4 + 25 = 29

5 + 20 = 25, which is the smallest sum and therefore the correct answer.