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 .

Factoring is basically removing quantities from an equation.  For instance  has an x in each term. Therefor the fully factored form would have the most amount of x's removed from each term and would be . You cannot take out more than is present in the smallest term (smallest in the sense of what you are trying to factor out). Here we could not remove more than one x since 2x was limiting. For Quadratic equations such as  we have a little easier time of factoring because we know that it will break down or factor into two binomials or two equations with two terms each. To get started recognize that since the "a" term is 1 , our first term in each binomial will just be x. Next you have to come up with two numbers that will multiply to equal the c term (the last one, 8), but add or subtract to give the middle "b" term. This comes from the fact the if you foiled your choices of your two factors the numbers must add or subtract to equal the middle term and multiply to equal the last term. Let's begin:

Start by writing two parentheses (    ) (    )

Since our "a" term is 1, the first term in each binomial will be x!

Now what numbers multiply to give the product of 8? 8 and 1. 2 and 4. Good. Which of these (if any because not every quadratic is factorable) are able to add to give the middle term of 6? Of course it is 2 and 4. So start by assuming the two factors are:

  Choose an addition sign because there is no way that 2 and 4 can subtract to equal a positive 6. Now us the FOIL method to check to see if this does become  when combined.

multply the first x by everything in the second parenthesis:

now mulitply 4 by every term in the second parenthesis and combine all like terms:

So  are the factors of . This step by step method makes the trial and error nature of this type of factoring a little more precise. Remember, if your "a" term is one then the first term in each of your factors are x. Then just think.. "what two numbers multiply to give me the last term of my quadratic I am factoring, but also add or subtract to give the middle term.

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into  quadratic form, but this is already done.

First, create two blank binomials.

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of  are  and , we know that

Next, factor out our constant, ignoring the sign for now. The factors of  are either  and  or  and . We must select those factors which have either a difference or a sum equal to the value of  in our trinomial. In this case,  and  cannot sum or difference to , but  and  can. Now, we can add in our missing values:

One last step remains. We must check our signs. Since  is negative in our trinomial, one and only one of our two binomials must have a negative sign. To figure out which, check the sign of  in our trinomial. Since  is negative, the larger of the two numbers in our binomial must be negative.

Thus, our two binomial factors are  and .

To factor a trinomial without using the quadratic equation, a few basic steps can be taken. The first step is always to rearrange our trinomial into  quadratic form, but this is already done.

First, create two blank binomials.

Start by factoring our first term back into the first term of each biniomial. Since the only reasonable roots of  are  and , we know that

Next, factor out our constant, ignoring the sign for now. The factors of  are either  and ,  and , or  and , We must select those factors which have either a difference or a sum equal to the value of  in our trinomial. In this case, neither  and  nor  and  can sum to , but  and  can. Now, we can add in our missing values:

One last step remains. We must check our signs. Since  is positive in our trinomial, Either both signs are negative or both are positive. To figure out which, check the sign of  in our trinomial. Since  is negative, both signs in our binomials must be negative.

Thus, our two binomial factors are  and .