Simplify the base of 9 and 27 in order to have a common base.

(3^x)(9)=27²

= (3^x)(3²)=(3³)²

=(3^x+2)=3⁶

Therefore:

x+2=6

x=4

The terms of \(\displaystyle 4x^{4}+ 36 x^{2}\) have \(\displaystyle 4x^{2}\) as their greatest common factor, so

\(\displaystyle 4x^{4}+ 36 x^{2} = 4x^{2} (x^{2}+9)\)

\(\displaystyle x^{2}+ 9\) is a prime polynomial. 

Of the five choices, only \(\displaystyle x^{2}+ 9\) is a factor.

The easiest way to approach this problem is to break everything into exponents. \(\displaystyle 8^{3}\) is equal to \(\displaystyle 2^{9}\) and 27 is equal to \(\displaystyle 3^{3}\). Therefore, the expression can be broken down into \(\displaystyle \frac{2^{9} \times 3}{2^{6}\times 3^{3}}\). When you cancel out all the terms, you get \(\displaystyle \frac{2^{3}}{3^{2}}\), which equals \(\displaystyle \frac{8}{9}\).

\(\displaystyle \sqrt{(45)(9)+(27)(36)}\)

When simplifying a square root, consider the factors of each of its component parts:

\(\displaystyle \sqrt{(3^2)\times5\times(3^{2})+(3^{3})(2^2)(3^2)}\)

Combine like terms:

\(\displaystyle \sqrt{5(3^4)+(2^2)(3^5)}\)

Remove the common factor, \(\displaystyle 3^4\):

\(\displaystyle \sqrt{(3^4)\times(5+3\times(2^2))}\)

Pull the \(\displaystyle \sqrt{3^4}\) outside of the equation as \(\displaystyle 3^2\):

\(\displaystyle (3^2)\sqrt{5+12}=9\sqrt{17}\)                       

\(\displaystyle \sqrt{(16)(8)+(32)(20)}\)

First, break down the components of the square root:

\(\displaystyle \sqrt{(2^{4})(2^{3})+(2^{5})(2^{2})\times 5}\)

Combine like terms. Remember, when multiplying exponents, add them together:

\(\displaystyle \sqrt{(2^{7})+(2^{7})\times5}\)

Factor out the common factor of \(\displaystyle 2^7\):

\(\displaystyle \sqrt{(2^{7})(1+5)}\)

\(\displaystyle \sqrt{(2^7)\times6}\)

Factor the \(\displaystyle 6\):

\(\displaystyle \sqrt{(2^7)\times2\times3}\)

Combine the factored \(\displaystyle 2\) with the \(\displaystyle 2^7\):

\(\displaystyle \sqrt{(2^{8})\times3}\)

Now, you can pull \(\displaystyle \sqrt{2^8}\) out from underneath the square root sign as \(\displaystyle 2^4\):

\(\displaystyle 2^4\sqrt{3}\)

\(\displaystyle \sqrt{(27)(45)(125)}\)

First, break down the component parts of the square root:

\(\displaystyle \sqrt{(3^{3})(5\times 3^{2})(5^{3})}\)

Combine like terms in a way that will let you pull some of them out from underneath the square root symbol:

\(\displaystyle \sqrt{(5^{4})(3^4)(3)}\)

Pull out the terms with even exponents and simplify:

\(\displaystyle (5^{2})(3^{2})\sqrt{3}=225\sqrt{3}\)

To find an equivalency we must rationalize the denominator.

To rationalize the denominator multiply the numerator and denominator by the denominator.

\(\displaystyle \frac{12}{\sqrt{18}}*\frac{\sqrt{18}}{\sqrt{18}}=\) 

\(\displaystyle \frac{12*\sqrt{18}}{18}=\)

Factor out 6,

 \(\displaystyle \frac{2*\sqrt{18}}{3}=\)

Extract perfect square 9 from the square root of 18.

\(\displaystyle \sqrt{18}=3*\sqrt{2}\)

\(\displaystyle \frac{2*3*\sqrt{2}}{3}=\) 

\(\displaystyle 2*\sqrt{2}\)