The first step is to divide this isosceles triangle into 2 right triangles, making it easier to solve:

The equation for area is 

We already know the base, so we need to solve for height to get the area. 

Then we plug in all values for the equation:

The first step toward finding the area is to divide this isosceles triangle into two right triangles:

Trigonometric ratios can be used to find both the height and the base, which are needed to calculate area:

With both of those values calculated, we can now calculate the area of the triangle:

C is given as 90°.

A is given as 22.620°

a is given as 5

b is given as 12

Therefore...

All angles of a triangle add up to equal 180°.

Synthetic division is a short cut for doing long division of polynomials and it can only be used when divifing by divisors of the form . The result or quoitient of such a division will either divide evenly or have a remainder. If there is no remainder, then the "" is said to be a factor of the polynomial. The polynomial must be in standard form (descending degree) and if a degree is skipped such as  it must be accounted for by a "place holder". 

___  __ __ __

    __ __ ___ 

where  is the remainder.  

While doing the long division we add vertically and we multiply diagonally by k. The empty lines represent places we put the sums and products. Notice that after the first term in the top row there is a 0; this is the place holder. This is because the degrees in the polynomial skipped. When the new coefficients have been found always rewrite starting with one order lower than the highest degree of the original polynomial.

Use synthetic division to verify each factor of the form . Lets start with .

Two goes into 6 three times resulting in:

_____________________

From here we see  will give you a remainder of zero and is therefore a factor of the polynomial .

Recall that dividing a polynomial by  does not always result in a pefect division (remainder of 0). Sometimes there is a remainder just like in normal division. When there is a remainder, we write the answer in a certain way.

For example 

 where the divisor is , the quotient or answer is , the remainder is , and the dividend is .

Even though we have variables here, this is the same as noting that  with a remainder of .

And how do we check to know if we have the right answer? We multiply  and add 3 to get 15, our dividend. The same method is used for synthetic division.

Thus, for our problem:

 ,

we must first multiply the divisor by the quotient using the foil method (first multiplying everything in the divisor by x and then everything in the divisor by 3) 

= 

now we just add the remainder which is 1 to yield  which matches the original dividend and is therefore our answer!

To determine if  is a root of the function given, you can use synthetic division to see if it goes in evenly. To set up the division problem, set up the coefficients of the function and then set 1 outside. Bring down the 1 (of the coefficients. Then multiply that by the  being divided in. Combine the result of that  with the next coefficient , which is . Then, multiply that by . Combine that result  with the next coefficient , which gives you . Multiply that by , which gives you . Combine that with the last coefficient , whcih gives you . Since this is not , you have a remainder, which means that  does not go in evenly to this function and is not a root.

Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

We multiply what's below the line by  and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

 with remainder 

This can be rewritten as

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

We multiply what's below the line by  and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

 with remainder 

This can be rewritten as:

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

We mulitply what's below the line by 1 and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder. 

 with remainder 

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.

Our first step is to list the coefficients of the polynomials in descending order and carry down the first coefficient.

Remember to place a  when there isn't a coefficient given.

We multiply what's below the line by  and place the product on top of the line. We find the sum of this number with the next coefficient and place the sum below the line. We keep repeating these steps until we've reached the last coefficients.

To write the answer, we use the numbers below the line as our new coefficients. The last number is our remainder.

 with remainder 

This can be rewritten as:

Keep in mind: the highest degree of our new polynomial will always be one less than the degree of the original polynomial.