# Factoring Part 3

The case we did not cover in my two previous hubs on factoring is a trinomial with leading coefficient 1 in which the middle term is either positive or negative and the last term is negative. This is the most challenging of the three because there are more sign possibilities with the factors. This means that you cannot count on both numbers being positive or both being negative. In fact, one number must be positive and one must be negative. The question is WHO GETS THE POSITIVE AND WHO GETS THE NEGATIVE. Now that all sounds complicated but i can greatly simplify things for you by providing the following statement: the sign when adding a positive and a negative is the SIGN OF THE LARGER NUMBER.

Let me demonstrate. Suppose we want to factor x^2 + 3x - 18. The possible factorizations of 18 are 1*18, 2*9, and 3*6. One of these numbers must be positive and the other must be negative because we want them to multiply and produce a negative 18. But when we add them we would like positive 3 to be the result. According to what i said above THE LARGER NUMBER SHOULD BE POSITIVE. Therefore our possibilities are really -1*18, -2*9, and -3*6. The last pairing here will give us what we want. -3*6 = -18 and -3+6 = 3. We conclude that x^2 + 3x - 18 = (x-3)(x+6).

Let's try another one. How about x^2 - 4x - 5. In this particular case there are only two numbers which multiply and equal 5, those being 1 and 5. Here we want THE LARGER NUMBER TO BE NEGATIVE because we ultimately want -4 after adding them. So quickly we see that 1 and -5 will do the job and we arrive at the answer (x+1)(x-5). The two examples illustrate that in this situation if the middle number is positive we stick a positive on the larger number and if the middle number is negative we stick a negative on the larger number. Not too bad then is it? With this in mind the two trinomials x^2 - 1x - 30 and x^2 + 5x - 14 become quickly factorable. -6*5=-30 and -6+5=-1(larger number must be negative) are satisfactory equations for the first problem as are 7*-2=-14 and 7+-2=5(larger number must be positive) for the second problem. (x-6)(x+5) and (x+7)(x-2) are then the factored trinomials. Try x^2 + 4x - 96 and x^2 - 6x - 27 for practice.

## Comments

No comments yet.