The greatest common factor (GCF) refers to the largest factor for each number that is shared by all the numbers of a given set, so in this case, it is the greatest factor of 15 and 75. The best way to approach these problems is to factorize each number and calculate the GCF from the multiplication of these numbers. 15 can be factorized into 3 and 5. 75 can be factorized into 3, 5, and 5. Now, identify the factors they have in common. Both 15 and 75 share one 3 and one 5, so multiplied together, the GCF = 3 * 5 = 15.

$\displaystyle \frac{4x^{5}y^{3}z}{12x^{3}y^{6}z^{2}}=\frac{x^{2}}{3y^{3}z}$

First, let's simplify $\displaystyle \frac{4}{12}$. The greatest common factor of 4 and 12 is 4. 4 divided by 4 is 1 and 12 divided by 4 is 3. Therefore $\displaystyle \frac{4}{12}=\frac{1}{3}$.

To simply fractions with exponents, subtract the exponent in the numerator from the exponent in the denominator. That leaves us with $\displaystyle \frac{1}{3}x^{2}y^{-3}z^{-1}$ or $\displaystyle \frac{x^{2}}{3y^{3}z}$

$\displaystyle \frac{32}{24}=1.33$, so the answer choice needs to be something that is not equal to $\displaystyle 1.33$ in order to be correct.

$\displaystyle \frac{16}{12}=1.33$

$\displaystyle \frac{224}{168}=1.33$

$\displaystyle \frac{4}{3}=1.33$

$\displaystyle \frac{160}{96}=1.67$

$\displaystyle y=\frac{x^2-\sqrt3x+\sqrt5x-\sqrt{15}}{x-\sqrt3}=\frac{(x-\sqrt3)(x+\sqrt5)}{x-\sqrt3}=x+\sqrt5$

The root occurs where $\displaystyle y=0$. So we substitute 0 for $\displaystyle y$.

$\displaystyle y=x+\sqrt{5}$

$\displaystyle 0=x+\sqrt{5}$

This means that the root is at $\displaystyle x=-\sqrt5$.

Find the common factors of the numerator and denominator.  They both share factors of 2,4, and 8.  For simplicity, factor out an 8 from both terms and simplify.

$\displaystyle \frac{32}{88}=\frac{8\times 4}{8\times 11}=\frac{4}{11}$

Remember that when you divide a fraction by a fraction, that is the same as multiplying the fraction in the numerator by the reciprocal of the fraction in the denominator. 

In other words,

$\displaystyle \frac{\frac{a^2}{b}}{\frac{a}{b}} = \frac{a^2}{b}\cdot\frac{b}{a}=\frac{a^2b}{ab}$

Simplifying this final fraction gives us our correct answer, $\displaystyle a$.

To solve for $\displaystyle x$, simplify the fraction. In order to do this, recall that dividing by a fraction is the same as multiplying by its reciprocal. Therefore, rewrite the equation as follows.

$\displaystyle \frac{\frac{x-2}{4^2}}{\frac{x^2-4}{2}}=1\Rightarrow \frac{x-2}{4^2}\times \frac{2}{x^2-4}=1$

Now, simplify the first fraction by calculating four squared.

$\displaystyle \frac{x-2}{16}\times \frac{2}{x^2-4}=1$

From here, factor the denominator of the second fraction.

$\displaystyle \frac{x-2}{16}\times \frac{2}{(x-2)(x+2)}=1$

Next, factor the 16.

$\displaystyle \frac{x-2}{2(8)}\times \frac{2}{(x-2)(x+2)}=1$

From here, cancel out like terms that are in both the numerator and denominator. In this particular case that includes (x-2) and 2.

$\displaystyle \frac{1}{8(x+2)}=1$

Now, distribute the eight.

$\displaystyle \frac{1}{8x+16}=1$

Next, multiply both sides by the denominator.

$\displaystyle (8x+16)\times \frac{1}{8x+16}=1\times (8x+16)$

The (8x+16) cancels out and leaves the following equation.

$\displaystyle 1=8x+16$

Now to solve for $\displaystyle x$ perform opposite operations to move all numerical values to one side of the equation leaving $\displaystyle x$ by itself on the other side of the equation.

$\displaystyle \\1=8x+16 \\-15=8x \\\\\frac{-15}{8}=x$

Let us simplify $\displaystyle \frac{6}{45}$:

$\displaystyle \frac{6}{45}=\frac{3\times 2}{3\times 15}=\frac{2}{15}$

We can get alternate forms of the same fraction by multiplying the denominator and the numerator by the same number:

$\displaystyle \frac{2\times 2}{15\times 2}=\frac{4}{30}$

$\displaystyle \frac{2\times 1.5}{15\times 1.5}=\frac{3}{22.5}$

Now let's look at $\displaystyle \frac{12}{89}$:

$\displaystyle 6 \times 2 = 12$, but $\displaystyle 45 \times 2 = 90$.

Therefore, $\displaystyle \frac{12}{89}$ is the correct answer, as it is not equivalent to $\displaystyle \frac{6}{45}$.