If you don't already have the pattern memorized, use FOIL. It's best to write out the parentheses twice (as below) to avoid mistakes:

\(\displaystyle (3n+5)^2\)

\(\displaystyle (3n+5)(3n+5)\)

\(\displaystyle 9n^2+15n+15n+25\)

\(\displaystyle 9n^2+30n+25\)

x² = 36 -> it is important to remember that this leads to two answers. 

x = 6 or x = -6. 

  If x = 6: A = B.

  If x = -6: A < B.

Thus the relationship cannot be determined from the information given.

We can use Heron's formula to find the area of the triangle. We can let a = 6, b = 10, and c = 12.

In order to find s, we need to find one half of the perimeter. The perimeter is the sum of the lengths of the sides of the triangle.

Perimeter = a + b + c = 6 + 10 + 12 = 28

In order to find s, we must multiply the perimeter by one-half, which would give us (1/2)(28), or 14.

Now that we have a, b, c, and s, we can calculate the area using Heron's formula. 

\(\displaystyle \sqrt[3]{-64x^{6}y^{12}}\)

Look for perfect cubes within each term. This will allow us to factor out of the radical.

\(\displaystyle \sqrt[3]{-64}*\sqrt[3]{x^6}*\sqrt[3]{y^{12}}\)

\(\displaystyle \sqrt[3]{-4*-4*-4}*\sqrt[3]{x^2*x^2*x^2}*\sqrt[3]{y^4*y^4*y^4}\)

Simplify.

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

Use the distributive property for radicals. 

\(\displaystyle \sqrt{5}(2\sqrt{3}+\sqrt{12})\)

Multiply all terms by \(\displaystyle \sqrt{5}\).

\(\displaystyle (\sqrt{5}*2\sqrt{3})+(\sqrt{5}*\sqrt{12})\)

Combine terms under radicals.

\(\displaystyle 2\sqrt{3*5}+\sqrt{12*5}\)

\(\displaystyle 2\sqrt{15}+\sqrt{60}\)

Look for perfect square factors under each radical. \(\displaystyle 60\) has a perfect square of \(\displaystyle 4\). The \(\displaystyle 4\) can be factored out.

\(\displaystyle 2\sqrt{15}+\sqrt{4*15}\)

\(\displaystyle 2\sqrt{15}+2\sqrt{15}\)

Since both radicals are the same, we can add them.

\(\displaystyle 4\sqrt{15}\)

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

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:

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

The complex conjugate of a complex number \(\displaystyle a+bi\) is \(\displaystyle a - bi\), so \(\displaystyle \frac{3}{2} + \frac{1}{4} i\) has \(\displaystyle \frac{3}{2} - \frac{1}{4} i\) as its complex conjugate. Subtract the latter from the former:

\(\displaystyle \left (\frac{3}{2} + \frac{1}{4} i \right ) - \left ( \frac{3}{2} - \frac{1}{4} i \right )\)

\(\displaystyle = \frac{3}{2} + \frac{1}{4} i - \frac{3}{2}+ \frac{1}{4} i\)

\(\displaystyle = \frac{3}{2}- \frac{3}{2} + \frac{1}{4} i + \frac{1}{4} i\)

\(\displaystyle = \frac{1}{2} i\)

The complex conjugate of a complex number \(\displaystyle a - bi\) is \(\displaystyle a+ bi\). Therefore, the complex conjugate of \(\displaystyle 17 - i \sqrt{17}\) is \(\displaystyle 17 + i \sqrt{17}\); subtract the latter from the former by subtracting real parts and subtracting imaginary parts, as follows:

\(\displaystyle (17 - i \sqrt{17})- (17 + i \sqrt{17})\)

\(\displaystyle = 17 - i \sqrt{17} - 17 - i \sqrt{17}\)

\(\displaystyle = 17 - 17 - i \sqrt{17} - i \sqrt{17}\)

\(\displaystyle = - 2 i \sqrt{17}\)