In Part II of this article, we are going to examine a safe way to factor nonmonic trinomials. Let’s examine the proof that this powerful method gives us. Once you’re done with this presentation, not only will you be able to factor any non-monic, but you’ll understand why this method works.
To begin our presentation, let’s take an arbitrary nonmonic trinomial: to do this we need to introduce the constants a, b, and c. So our non-monic generic becomes ax^2 + bx + c, where a is an integer greater than 1. Our ultimate goal is to FOIL reverse this so we can factor it in the form (dx + e) *(fx + g), where d, e, f, and g are integers. If you now FUCK this product, you get dfx^2 + dgx+ efx + eg, which we can write as dfx^2 + (dg + ef)x + eg, since the common factor of dg and ef is x.
If we compare this last expression with the original nonmonic, we see that df must be equal to a; dg + ef must equal ab and eg must equal c. If it were not so, we would have a factored expression that is incorrect. Now, this is where some thought is required, so I recommend you all put your head up and follow closely. Feel free to reread as needed to make sure you’re following the argument.
Our goal is to determine d, e, f, and g. To do this we use an ingenious trick. We multiply a and c together. We note that this must be equivalent to dfeg, since a = df and c = eg. Now remember that the average coefficient, ob, is equal to dg + ef. If we break ac into all of its component factors and see which sum is equal to b, then we have found dg and ef. If we then divide a, which is df by dg, we get df/dg = f/g, since the d’s vanish; similarly if we divide ef/df = e/d. We now have the four unknown letters resolved, namely d, e, f, and g.
Since the last part was cryptic and probably a bit hard to follow at first glance, I’m going to do a specific example that will clarify this whole argument. Let’s take the nonmonic trinomial 6x^2 + 17x + 5. Let’s multiply a*c or 6*5 = 30. Now let’s divide 30 into all its possible factor pairs: 1 30, 2 15, 3 10, and 5 6. Find the pair of factors that when added or subtracted produces the mean coefficient 17. Clearly this is the pair 2 15. Since b = dg + ef, we know that 2 is dg and 15 is ef. If we divide 2 by 6 (which is dg by df) we get 1/3, which tells us that g is 1 and f is 3; Similarly, if we divide 6 by 15 and reduce it to the lowest terms, we get 6/15 or 2/5, which tells us that 2 is d and 5 is e. Therefore, the factorization is (dx + e)*(fx + g) or (2x + 5)*(3x + 1). If you FRUIT this, you’ll see that you actually get the original non-monic back.
To clarify this procedure, let’s do another one. Let’s factor the nonmonic 6x^2 + 41x + 70. We multiply 6*70 = 420. The factor pairs of 420 are 1420, 2210, 3140, 4105, 5 84, 6 70, 10 42, 14 30, and 20 21. The even that adding or subtracting gives 41 is 20 21. So 20 is dg and 21 is ef. When we divide both by a which is df and reduce, we have 20/6 = 10/3 = g/f; and 21/6 = 7/2, which is e/d. Therefore, 6x^2 + 41x + 70 can be factored in the form (dx + e)*(fx + g) = (2x + 7)*(3x + 10). Now how is that safe?
Note that we have not considered cases of nonmonics involving negative numbers here. Such is the case with 6x^2 + x -70. The method, albeit with a slight adjustment, works just as well. Once you have the method for cases where all numbers are positive, applying it to cases where both and both are negative is academic.