Therefore:

This is a quadratic equation in standard form, so first we need to factor .

This can be factored out as 

where .

By trial and error we find that , so

can be expressed as

.

Set each linear binomial equal to 0 and solve separately:

The solution set is .

The area of a right triangle with legs of length  and  is 

.

Substitute  and  for  and  and 600 for , then solve for :

We can now factor the quadratic expression:

Set each linear binomial to 0 and solve to get possible solutions:

Since  must be positive, we throw out the negative solution. 

.

This is a factoring problem so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract  from both sides to get 

Think of the equation in this format to help with the following explanation.

We must then factor to find the solutions for . To do this we must make a factor tree of  which is 28 in this case to find the possible solutions. The possible numbers are , , .

Since  is positive we know that our factoring will produce two positive numbers.

We then use addition with the factoring tree to find the numbers that add together to equal . So ,  , and 

Success! 14 plus 2 equals . We then plug our numbers into the factored form of 

We know that anything multiplied by 0 is equal to 0 so we plug in the numbers for  which make each equation equal to 0 so in this case .

The center is located at (3,4) which means the standard equation of a circle which is:

becomes

which equals to

Change division into multiplication by the reciprocal which gives us the following

Now 

this results in the following:

Simplification gives us

  which equals

This expression involves the difference of two cubic terms. To factor an expression in this format, we can use a special formula.

Before we can use this formula, we need to manipulate our original expression to identify and .

Comparing this with the formula,  and . Now we can use the formula to factor.

This problem involves the difference of two cubic terms. We need to use a special factoring formula that will allow us to factor this equation.

But before we can use this formula, we need to manipulate  to make it more similar to the left hand side of the special formula. We do this by making the coefficients (343 and 64) part of the cubic power.

Comparing this with ,  and .

Plug these into the formula.

Begin by factoring out a 2:

Then, we recognize that the trinomial can be factored into two terms, each beginning with :

Since the last term is negative, the signs of the two terms are going to be opposite (i.e. one positive and one negative):

Finally, we need two numbers whose product is negative thirty-five and whose sum is positive two.  The numbers  and  fit this description.  So, the factored trinomial is:

You can factor this trinomial by breaking it up into two binomials that lead with :

You will fill in the binomials by finding two factors of 36 that add up to 5. This is achieved with positive 9 and negative 4:

You can then set each of the two binomials equal to 0 and solve for :