** ** As a Middle School teacher who has taught algebra in the
Middle School, I have found that students have a most difficult time in
factoring equations of the form

Let us take a look at a simple example demonstrating the method to which they are now exposed and see for ourselves how the process could demoralise young students.

__Example :__

Using the method recommended by most of the textbooks, students would proceed as follows:

__Step 1 __

Open two pairs of parentheses and find 2 factors of 8

Thus, (4x )(2x ) = 0 might be a way of proceeding.

__Step 2__

Find 2 factors of ^{-}6 and include these in our
parentheses.

This gives (4x - 3)(2x + 2) = 0

__Step 3__

Check to see if the middle term is correct. This is done by finding the product of the two innermost terms and the two outermost terms in the brackets and then adding them to see if they are the same as the secon term in the quadratic, i.e. 13x in this case. If we check the factors above, we obtain:

Product of innermost terms = ^{-}6x

Product of outermost terms = 8x

Sum = 2x

Since this sum is not equal to 13, then we have not factored correctly.
The student must now experiment with other combinations of factors of ** a**
and **c** until the desired situation is obtained.

In the initial stages, this can be very frustrating to the young student. In the case of middle school students, this approach can have very serious effects on the motivation of students to do higher math. I have experimented with this type of problem and finally came up with the following procedure. This approach works every time and has helped to provide a "comfort zone" for my students where they can study the consistency of factoring quadratics from a position of strength.

Let us consider the same problem i.e.

Solve 8x^{2} + 13x - 6 = 0

__Step 1__:

__Determine factorability.__

__ Find the product of __**a **and **c** and identify factors
of this product which add up to **b**.

In this case **ac** = 8 x ^{-}6 = ^{-}48. The
factors of this product which add up to 13 are 16 and ^{-}3

__Step 2__:

Open 2 pairs of parentheses and find factors of **a **which are also
individually factors of at least 1 of the factors of **ac **found in Step
1

In this case we would obtain:

( 8x )(x )

Almost immediately it becomes clear to us that we cannot use 2 and 4,
because we cannot divide ^{-}3 by 2. The factors 8 and 1 are the ones
we want since 8 divides 16 and 1 divides ^{-}3.

__Step 3:__

__ We now need to arrange for our products of innermost and outermost
terms to have the same coefficients as the factors of__** ac **found in Step
1.

In this case, what can we multiply 8 by to give 16? Clearly, the answer
is 2. This answer now becomes the second outer term in the pair of parentheses
in Step 2. Similarly, we can find the second innermost term by identifying the
number we must multiply 1 by to give ^{-}3. Clearly, the answer is
^{-}3 which is in fact the second innermost term. Our complete answer
is:

(8x - 3)(x + 2) = 0

__Step 4:__

The final step is to find the product of the 2 last terms in the bracket
and ensure that this product is equal to **c **in the original equation.

In this case, we have satisfied the conditions of Step 4.

This approach works every time and it eliminates all the trauma which goes with a "guess and check" method especially where young students are concerned.

Another example should make the approach clearer.

__Example:__

__ Solve 6x__^{2} - x -12 = 0

**ab = **^{-}72; Factors of ^{-}72 which will add to
^{-}1 **(b) ** are ^{-}9 and 8, so the quadratic is
factorable.

The factors of 6 where each will divide at least one of the factors of
^{-}9 and 8 are 2 and 3. (2 will divide 8 and 3 will divide
^{-}9).

Thus, (2x )(3x ) = 0

To complete our factorization we need to consider 8x and ^{-}9x as
the product of our outer and inner terms respectively.

And so (2x - 3)(3x + 4) = 0

The product of our end terms is ^{ -}12, confirming that we have
factored correctly.