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 .

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.

 ---> 

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 . Note that this can also be written as , and if you graph this, you get a result identical to .

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. Note that this is a special trinomial: , and thus only  and  are present to be manipulated.

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 only 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, only  and  sum to , which we use since  is not present.

One last step remains. We must check our signs. Since  is negative in our trinomial, one sign is positive and one is negative.

Thus, our two binomial factors are  and .

This is called a difference of squares. If you see a trinomial in the form , the roots are always  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. It isn't ideal to work with , but  in this equation cannot be reduced.

 ---> 

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  and . Note that these do not normally sum or difference to , but the presence of  as a term means one term will be doubled in value when creating . This means we either get  and  (no good), or  and  (good). Remember to add the number we want to double inside the binomial opposite , so it multiplies correctly. Now, we can add in our missing values:

One last step remains. We must check our signs. Since  is negative in our trinomial, one sign is positive and one is negative. To figure out which, check the sign of  in our trinomial. Since  is positive, the bigger of our two terms after any multipliers are completed is the positive value. Since our terms after mutiplying are  and , the  must be the positive term.

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. It isn't ideal to have , but we cannot reduce the constants further.

First, create two blank binomials.

Start by factoring our first term back into the first term of each biniomial. There are two valid factorings of . We could have  or , and there's not really a valid system for figuring out which is correct (unless you use the quadratic formula). As a loose rule of thumb, if  is near in value to , it's more likely (though by no means necessary) that the factors of  are also close in value. So, let's try .

Next, factor out our constant , ignoring the sign for now. The factors of  are either  and , or  and , but the presence of  and  as terms means one term will be doubled in value and the other will be tripled when creating . So, for  and  we either produce  and  (no good), or  and  (no good) for . If we try  and  with  and , we get either  and  (no good) or  and  (good!) Remember to add the numbers we want to use opposite the  term we want to combine them with , so it multiplies correctly. 

Now, we can add in our missing values:

One last step remains. We must check our signs. Since  is negative in our trinomial, one sign is positive and one is negative. To figure out which, check the sign of  in our trinomial. Since  is positive, the bigger of our two terms after any multipliers are completed is the positive value. Since our terms after mutiplying are  and , the number that becomes  must be the positive term.

Thus, our two binomial factors are  and .

Through factoring, the sum of the two roots must equal 2, and the product of the two roots must equal . , and 3 satisfy both of these criteria.