Given the equation 

Realizing that the domain is restricted by the denominator, meaning that the denominator can not be equal to 0. 

Set the denominator = to 0 and solve. 

First factoring  , 

Zero-product Property, setting both quantities equal to 0 and solve. 

So when x is 6 or -1, our denominator will be 0. Meaning those would be our domain restrictions. 

To factor this, we need to find a pair of numbers that multiplies to 6 and sums to 5. The numbers 2 and 3 work. (2 * 3 = 6 and 2 + 3 = 5)

So (x + 2)(x + 3) = 0

x = –2 or x = –3

To find the domain, find all areas of the number line where the fraction is defined.

because the denominator of a fraction must be nonzero.

Factor by finding two numbers that sum to -2 and multiply to 1.  These numbers are -1 and -1.

To factor the polynomial, we need the numbers that multiply to –12 and add to +1. This leads us to –3 and +4. We solve the polynomial by setting it equal to 0.

So either x = 3 or x = –4 will make the product equal to 0.

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 128 from both sides to get .

We then notice that all four numbers are divisible by four, so we can simplify the expression to .

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

We must then factor to find the solutions for x. To do this we must make a factor tree of c (which is 32 in this case) to find the possible solutions. The possible numbers are 1 * 32, 2 * 16, and 4 * 8.

Since c is negative, we know that our factoring will produce a positive and negative number.

We then look at b to see if the greater number will be positive or negative. Since b is positive, we know that the greater number from our factoring tree will be positive. 

We then use addition and subtraction with the factoring tree to find the numbers that add together to equal b. Remember that the greater number is positive and the lesser number is negative in this example.

Positive 8 and negative 4 equal b. 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 x which make each equation equal to 0. In this case .

The quadratic can be solved as . Setting each factor to zero yields the answers.

When we attempt to factor a quadratic, we must first look for the factored numbers. When quadratics are expressed as  the factored numbers are  and . Since , we know the factors for 1 are 1 and 1. So we know the terms will be

Looking at our constant, , we see a positive 6. So 6 factors into either 2 and 3  or 1 and 6 (since  and ). Since our constant is a positive number, we know that our factors are either both positive, or both negative. (Note: you should know that 2 negative numbers multiplied becomes a positive number).

So to figure out what we must use we look at the  part of the quadratic. We are looking for 2 numbers which add up to our . So, 1 and 6 do not work, since . But, 2 and 3 are perfect since .

But, since our  is a negative 5, we know we must use negative numbers in our factored expression. Thus, our factoring must become

or

The expression can be factored by finding terms that multiply back to the original expression. The easiest way is to find two numbers that add to the middle term  as well as multiply to the last term  The numbers that satisfy both of these conditions are  and , so the answer is .

Determine the signs of the binomials.  Because the quadratic has a negative middle term and a positive end term, the signs will be both negative.

To factor, find the roots of  that will either add or subtract to get a  coefficient.

The only possibility is the set .

Substitute the roots to the factored form .